Inequalities

An inequality controls a quantity even when its exact value is inconvenient or unnecessary. A useful proof must establish three things: the operations are valid on the stated domain, each bound goes in the required direction, and any claimed equality can actually occur.

This chapter builds on the ordered-field rules in Number Systems, Algorithms, and Recursion and the notation in Finite Sums and Summation Techniques. We work with real numbers and finite sums unless otherwise stated. Loh’s inequality notes provide a compact companion list of the classical results [1][1] P. S. Loh, “6. Inequalities,” 2023. Carnegie Mellon University, Putnam Seminar, Fall 2023; classical inequalities. https://www.math.cmu.edu/~ploh/docs/math/2023-295/06-inequalities.pdf.

Transform an inequality without changing its meaning

To prove $A\ge B$, it is often enough to show $A-B\ge0$. But moving between expressions requires conditions. For real $a,b$ and a positive integer $n$, keep the following rules explicit.

Operation	Condition and effect
Add the same number	Preserves order on all of $\mathbb R$
Multiply or divide by $c$	Preserves order if $c>0$; reverses it if $c<0$; division is undefined at $c=0$
Take reciprocals	Reverses order on each of $(-\infty,0)$ and $(0,\infty)$ separately
Raise to an odd power	Preserves strict order on all of $\mathbb R$
Raise to an even power	Preserves order on $[0,\infty)$; reverses it on $(-\infty,0]$
Take a positive integer root	Preserves order on its real domain; an even root requires a nonnegative input

For example, $-1>-2$ but $(-1)^2<(-2)^2$. Also, $-1<2$ does not imply $1/(-1)>1/2$: the reciprocal rule cannot cross zero. If $a,b\ge0$, then $a\le b$ is equivalent to $a^2\le b^2$; without that restriction, squaring can lose sign information.

Absolute-value bounds translate into intervals. For $r\ge0$,

$$|x|\le r\quad\Longleftrightarrow\quad -r\le x\le r.$$

The strict version $|x|<r$ gives the open interval $(-r,r)$ for $r>0$. If $r<0$, $|x|\le r$ has no real solution. Check these boundary cases before applying a general-looking rule.

Solve inequalities by signs and intervals

For a genuine quadratic $q(x)=ax^2+bx+c$, assume $a\ne0$ and let $\Delta=b^2-4ac$. If $a>0$, the sign pattern is:

Discriminant	Sign of the quadratic
$\Delta>0$, roots $r_1<r_2$	Positive outside the roots; negative between them; zero at the roots
$\Delta=0$, root $r$	Positive except at $r$, where it is zero
$\Delta<0$	Positive everywhere

If $a<0$, apply the table to $-q$ and reverse signs. If $a=0$, solve the resulting linear or constant inequality instead. For a non-strict inequality, include zeros where appropriate.

For example,

$$4x^2+6x+2=2(2x+1)(x+1).$$

The roots are $-1$ and $-1/2$, and the leading coefficient is positive. Therefore $4x^2+6x+2<0$ has solution $(-1,-1/2)$. Testing $x=-3/4$ inside the interval gives a negative value, which is a useful check on the selected region.

View the three quadratic sign patterns

The picture supports the sign table; the factorization or completed square gives the algebraic justification. Endpoints belong to a solution only if the inequality permits equality.

For rational expressions, denominator zeros are excluded even if cancellation seems possible. To solve

$$\frac{x-1}{x+2}\ge0,$$

split the number line at $-2$ and $1$. The quotient is positive on $(-\infty,-2)$ and $(1,\infty)$, negative on $(-2,1)$, and zero at $1$. Thus the solution is $(-\infty,-2)\cup[1,\infty)$. Multiplying by $x+2$ without first determining its sign would miss this structure.

Absolute values and the triangle inequality

TheoremTriangle inequality

For all real $a,b$,

$$|a+b|\le|a|+|b|.$$

Equality holds exactly when $ab\ge0$, including the cases where either number is zero.

Proof

Both sides are nonnegative, so comparing their squares is equivalent. Their squared difference is

$$\begin{aligned} &(|a|+|b|)^2-|a+b|^2\\ &\quad=2(|ab|-ab)\ge0. \end{aligned}$$

It vanishes exactly when $ab\ge0$.

In distance language, a direct journey is no longer than a journey through an intermediate point:

$$|a-c|\le|a-b|+|b-c|.$$

Apply the theorem to $(a-b)+(b-c)$. Equality holds exactly when $b$ lies between $a$ and $c$, endpoints included.

TheoremReverse triangle inequality

For real $a,b$,

$$\bigl||a|-|b|\bigr|\le|a-b|.$$

Equality holds exactly when $ab\ge0$.

Proof

Writing $a=(a-b)+b$ gives $|a|-|b|\le|a-b|$. Interchanging $a,b$ gives the other direction, so the absolute difference is bounded. For the equality condition, compare the squares of $|a-b|$ and $\bigl||a|-|b|\bigr|$; their difference is $2(|ab|-ab)$.

The reverse inequality says that changing a real input by at most $\delta$ changes its absolute value by at most $\delta$. This is useful for error estimates and later for distances and complex moduli.

For a finite sequence of real terms,

$$\left|\sum_{i=1}^{n}x_i\right|\le\sum_{i=1}^{n}|x_i|.$$

Repeated application of the two-term inequality, or induction on $n$, proves this. It includes the three-term version; for $n=0$, both empty sums are zero. Equality holds precisely when the nonzero terms all have the same sign: otherwise the positive and negative totals cancel partially.

Arithmetic and geometric means

TheoremTwo-term AM-GM

For $a,b\ge0$,

$$\sqrt{ab}\le\frac{a+b}{2},$$

with equality exactly when $a=b$.

Proof

The inequality $(\sqrt a-\sqrt b)^2\ge0$ gives $a+b\ge2\sqrt{ab}$, and the square vanishes exactly when $a=b$.

A related identity applies to all real $a,b$, with no positivity assumption:

$$a^2+b^2-2ab=(a-b)^2\ge0.$$

This is often the quickest way to create a bound from a difference of expressions.

TheoremFinite AM-GM

For $n\ge1$ and nonnegative $x_1,\ldots,x_n$,

$$\left(\prod_{i=1}^{n}x_i\right)^{1/n} \le\frac1n\sum_{i=1}^{n}x_i.$$

Equality holds exactly when all $x_i$ are equal.

Proof by doubling and then descending

Proof

First suppose all terms are positive. The two-term result is already known. If AM-GM holds for $m$ terms, split $2m$ terms into equal-sized groups with arithmetic means $A,B$. Each group’s product is at most the $m$th power of its mean, and $AB\le((A+B)/2)^2$. Multiplying proves the result for $2m$ terms. Starting at one term gives every power-of-two case.

Next suppose the result holds for $m+1$ terms. Given $m$ positive terms with mean $M$, append the extra term $M$. The new mean is still $M$, so

$$x_1\cdots x_m M\le M^{m+1}.$$

Cancel positive $M$ to obtain the $m$-term result. For any desired $n$, start with a power of two at least $n$ and descend finitely many times. Equality in the doubling argument requires equal entries within each group and equal group means; descending retains exactly the all-equal case.

Finally, if some terms are zero, the product is zero. Equality then holds only if the arithmetic mean is also zero, which for nonnegative terms means all terms are zero.

The harmonic-to-quadratic mean chain

For positive $x_1,\ldots,x_n$, define

$$H=\frac{n}{\sum_{i=1}^{n}1/x_i},$$ $$G=\left(\prod_{i=1}^{n}x_i\right)^{1/n},$$ $$A=\frac1n\sum_{i=1}^{n}x_i,$$ $$Q=\sqrt{\frac1n\sum_{i=1}^{n}x_i^2}.$$

Then

$$0<H\le G\le A\le Q.$$

The middle inequality is AM-GM. Applying AM-GM to the reciprocals and then taking reciprocals gives $H\le G$. For the last inequality,

$$Q^2-A^2=\frac1n\sum_{i=1}^{n}(x_i-A)^2\ge0.$$

Taking nonnegative square roots proves $A\le Q$. Equality in any of the three comparisons holds exactly when all inputs are equal. Strict positivity is needed for the harmonic mean; the other three means still make sense for nonnegative inputs.

For two equal-distance trips at positive speeds $v_1,v_2$, average speed is total distance divided by total time, giving

$$v_{\mathrm{avg}}=\frac{2}{1/v_1+1/v_2}.$$

At $60$ and $40$ km/h, this is $48$ km/h. Equal times instead give the arithmetic mean, $50$ km/h. The averaging rule depends on what is held equal.

Exploration: what the mean surfaces show

For nonnegative $x_1,x_2$, compare the surfaces $z=(x_1+x_2)/2$ and $z=\sqrt{x_1x_2}$.

The AM-GM proof shows which surface lies above the other. Equality occurs along $(x_1,x_2,z)=(t,t,t)$ for $t\ge0$. Fixing $x_2=c$ gives a vertical slice, not a horizontal one. Its two curves meet at $(x_1,z)=(c,c)$.

Thus changing the fixed second input from $5$ to $10$ moves the intersection to $(10,10)$ in that slice. For $c=0$, the geometric mean is zero and equality occurs only at $x_1=0$. A projected front view of a surface is not, by itself, the same construction as taking a slice.

Cauchy-Schwarz controls a sum of products

For real vectors $a=(a_1,\ldots,a_n)$ and $b=(b_1,\ldots,b_n)$, write

$$\begin{aligned} A&=\sum_{i=1}^{n}a_i^2,\\ B&=\sum_{i=1}^{n}b_i^2,\\ C&=\sum_{i=1}^{n}a_i b_i. \end{aligned}$$

TheoremCauchy-Schwarz

For $n\ge1$, $C^2\le AB$. Equivalently,

$$\left|\sum_{i=1}^{n}a_i b_i\right| \le\sqrt{\sum_{i=1}^{n}a_i^2}\sqrt{\sum_{i=1}^{n}b_i^2}.$$

Equality holds when one vector is zero, or when the two nonzero vectors are real scalar multiples of one another.

Proof

The Lagrange identity proved in Finite Sums states that

$$AB-C^2=\sum_{1\le i<j\le n}(a_i b_j-a_j b_i)^2.$$

The right side is nonnegative. It vanishes exactly when every $a_i b_j=a_j b_i$. If $b\ne0$, choose a coordinate $j$ with $b_j\ne0$; then every $a_i=(a_j/b_j)b_i$. If $b=0$, equality holds for every $a$. This handles zero coordinates without dividing by all of them.

For $n=2$, the difference is the single square $(a_1b_2-a_2b_1)^2$. For $n=3$, it is the sum over the three coordinate pairs. These are special cases of the same theorem, rather than separate geometric assumptions about dimensions.

An independent proof using a nonnegative quadratic

Proof

If $B=0$, then $b=0$ and the assertion holds. Otherwise, for every real $t$,

$$\begin{aligned} 0&\le\sum_{i=1}^{n}(a_i-tb_i)^2\\ &=A-2tC+t^2B. \end{aligned}$$

Choose $t=C/B$. The right side becomes $A-C^2/B$, proving $C^2\le AB$. Equality holds exactly when all $a_i-tb_i$ vanish. Equivalently, this quadratic has nonpositive discriminant; the leading coefficient must first be known to be positive.

Inner product, length, and the correct triangle connection

The dot product is $a\cdot b=C$, and the Euclidean norm is $\|a\|_2=\sqrt A$. Cauchy-Schwarz becomes

$$|a\cdot b|\le\|a\|_2\|b\|_2.$$

In the plane or three-dimensional space, when both vectors are nonzero, their angle satisfies $a\cdot b=\|a\|_2\|b\|_2\cos\theta$. The dot product is not the cosine itself. Equality in the absolute-value bound allows the same or opposite direction; if a vector is zero, the angle is undefined but the inequality still holds with equality.

To derive the Euclidean triangle inequality, expand the squared norm and bound the cross term:

$$\begin{aligned} \|a+b\|_2^2&=A+B+2C\\ &\le A+B+2\sqrt{AB}\\ &=(\sqrt A+\sqrt B)^2. \end{aligned}$$

Taking square roots gives $\|a+b\|_2\le\|a\|_2+\|b\|_2$, the $p=2$ Minkowski inequality. Equality requires one vector to be zero or the nonzero vectors to point in the same direction. Setting unused coordinates to zero now recovers the scalar triangle inequality. The expansion is essential; specializing Cauchy-Schwarz alone does not produce it.

A useful form for positive denominators

For real $u_i$ and positive $v_i$,

$$\sum_{i=1}^{n}\frac{u_i^2}{v_i} \ge\frac{(\sum_{i=1}^{n}u_i)^2}{\sum_{i=1}^{n}v_i}.$$

Apply Cauchy-Schwarz to the entries $u_i/\sqrt{v_i}$ and $\sqrt{v_i}$, then divide by the positive sum of denominators. Equality holds exactly when every ratio $u_i/v_i$ has the same value. This form is often called Engel’s form of Cauchy-Schwarz.

For positive $x_i$, putting $u_i=1$, $v_i=x_i$ yields

$$\sum_{i=1}^{n}\frac1{x_i}\ge\frac{n^2}{\sum_{i=1}^{n}x_i},$$

with equality exactly when all $x_i$ are equal.

Rearrangement: improve a pairing by swapping

Let $n\ge1$. Suppose $a_1\le\cdots\le a_n$ and $b_1\le\cdots\le b_n$ are real; positivity is not required. Let $\sigma$ be a permutation and put

$$S_\sigma=\sum_{i=1}^{n}a_i b_{\sigma(i)}.$$

TheoremRearrangement inequality

Pairing in opposite order minimizes the sum, while pairing in the same order maximizes it:

$$\sum_{i=1}^{n}a_i b_{n+1-i} \le S_\sigma\le\sum_{i=1}^{n}a_i b_i.$$

Proof

Suppose $i<j$ but the assigned values satisfy $b_{\sigma(i)}>b_{\sigma(j)}$. Swapping these two assignments changes the sum by

$$(a_j-a_i)(b_{\sigma(i)}-b_{\sigma(j)})\ge0.$$

Repeatedly removing adjacent inversions sorts the assigned $b$ values and reaches the same-order pairing, without decreasing the sum. Sorting in the opposite direction instead removes same-direction pairs without increasing the sum, giving the minimum.

Equality in the upper bound means there is no pair with $a_i<a_j$ but $b_{\sigma(i)}>b_{\sigma(j)}$; equality in the lower bound means there is no pair with $a_i<a_j$ and $b_{\sigma(i)}<b_{\sigma(j)}$. Ties allow zero-cost swaps. If both original sequences are strictly increasing, only the same-order permutation maximizes and only the reverse-order permutation minimizes. Equality does not generally require all entries to be identical.

For denominations $10,20,50,100$ and quantities $2,3,4,5$, the maximum total is $780$, from the same-order pairing; the minimum is $480$, from opposite order. Both are attained by explicit assignments.

Chebyshev’s sum inequality

For $n\ge1$, if two real sequences are sorted in the same direction, then

$$\frac1n\sum_{i=1}^{n}a_i b_i \ge\left(\frac1n\sum_{i=1}^{n}a_i\right) \left(\frac1n\sum_{i=1}^{n}b_i\right).$$

For opposite directions, reverse the inequality. The identity behind both statements is

$$\begin{aligned} &n\sum_{i=1}^{n}a_i b_i -\left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}b_i\right)\\ &\quad=\sum_{i<j}(a_i-a_j)(b_i-b_j). \end{aligned}$$

Expand the right side to verify it. Same-order pairs have nonnegative products; opposite-order pairs have nonpositive products. Equality holds exactly when at least one sequence is constant: if both are nonconstant and similarly ordered, their first and last entries already give a strictly positive term. The opposite-order argument is analogous.

This is a result about sums of ordered sequences. It is distinct from the probability inequality that bounds tail probabilities using variance. Loh’s earlier handout lists the sum inequality alongside rearrangement [2][2] P. S. Loh, “II. Inequalities,” 2003. June 18, 2003; notes on norms, rearrangement, and Chebyshev's sum inequality. https://www.math.cmu.edu/~ploh/docs/math/2-inequalities-solns.pdf.

Choose a method and check equality

Shape of the problem	A useful first attempt
A difference of quadratic expressions	Complete squares or factor the difference
A product with a fixed sum	AM-GM, with nonnegative inputs
Absolute values or accumulated errors	Triangle or reverse triangle inequality
A sum of products	Cauchy-Schwarz
Squared numerators over positive denominators	Engel’s form
A pairing that can be permuted	Rearrangement

If scaling every variable by $t>0$ multiplies both sides by the same factor $t^d$, the inequality is homogeneous of degree $d$. This permits normalization: for positive variables with sum $s$, dividing every variable by $s$ gives an equivalent sum-one problem. Check that the transformation preserves the desired inequality. Equality conditions suggest where an optimum might occur, but a lower or upper bound becomes an attained minimum or maximum only after an admissible equality case is exhibited.

Exercises

ExerciseSupply the missing sign conditions

For real variables, assess the implications $a>b\Rightarrow a/c>b/c$, $ac<bc\Rightarrow a<b$, $a<b\Rightarrow1/a>1/b$, $ac^2>bc^2\Rightarrow a>b$, and $a>b\Rightarrow a^n>b^n$ for positive integers $n$.

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Solution

The first two conclusions require $c>0$ for the displayed direction; $c<0$ reverses it, and division by zero is forbidden. In the second implication, $c=0$ makes the premise impossible. The reciprocal implication holds when $a,b$ have the same nonzero sign, but fails across zero. The fourth implication already forces $c\ne0$, so dividing by positive $c^2$ proves it. Odd powers preserve order on all reals; even powers require restricting to a range on which the power function has the appropriate monotonicity, such as $0\le b<a$ for the stated direction.

ExerciseCompare two negative quotients

Let $a>b>0$, $c<d<0$, and $f<0$. Prove

$$\frac{f}{a-c}>\frac{f}{b-d}.$$

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Solution

Both denominators are positive, and their difference is $(a-b)+(d-c)>0$. Taking reciprocals gives $1/(a-c)<1/(b-d)$. Multiplying by negative $f$ reverses the direction. The larger positive denominator makes the negative quotient closer to zero, hence larger.

ExerciseRepeated roots and negative leading coefficients

Solve $4x^2+4x+1<0$ and $-3x^2+x-6<0$. How does the first answer change if $<$ becomes $\le$?

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Solution

The first expression is $(2x+1)^2$, so the strict inequality has no real solution. With $\le0$, the only solution is $x=-1/2$. For the second, negate the polynomial and reverse the inequality. The resulting $3x^2-x+6$ has positive leading coefficient and discriminant $-71$, so it is positive for all real $x$. The original strict inequality therefore holds on all of $\mathbb R$.

ExerciseAllocate an error budget

Let $\varepsilon>0$. If $|x-a|<\varepsilon/4$ and $|y-b|<\varepsilon/6$, prove $|2x+3y-2a-3b|<\varepsilon$.

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Solution

Regroup first, then use the triangle inequality:

$$\begin{aligned} &|2(x-a)+3(y-b)|\\ &\quad\le2|x-a|+3|y-b|\\ &\quad<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. \end{aligned}$$

The denominators in the hypotheses were chosen so the two contributions fit the total error budget.

ExerciseBound both diagonals without discarding terms

For real $a,b,c,d$, prove

$$\begin{aligned} |a-c|+|b-d|&\le |a-b|+|b-c|\\ &\quad+|c-d|+|d-a|. \end{aligned}$$

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Solution

Call the right side $P$. Bound $|a-c|$ along each of the two routes, through $b$ and through $d$, then add the bounds to obtain $2|a-c|\le P$. Likewise, the routes from $b$ to $d$ through $a$ and through $c$ give $2|b-d|\le P$. Add these two inequalities and divide by two. One cannot remove a positive term from a previously obtained upper bound and assume the bound remains valid.

ExerciseMultiply two AM-GM bounds

For $a,b,c>0$, prove $(a+b+c)(a^2+b^2+c^2)\ge9abc$ and determine equality.

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Solution

AM-GM gives

$$a+b+c\ge3(abc)^{1/3},$$$$a^2+b^2+c^2\ge3(abc)^{2/3}.$$

All quantities are positive, so the bounds may be multiplied. Equality requires equality in both applications, exactly when $a=b=c$.

ExerciseFind an attained minimum

For $x,y,z>0$, minimize $x/y+y/z+z/x$.

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Solution

The three positive terms have product one, so AM-GM gives a lower bound of $3$. Equality requires all three ratios equal; their product then forces each ratio to be one. Taking $x=y=z$ attains the bound, so the minimum is $3$.

ExerciseA sum-one normalization

For positive $a,b,c$ with $a+b+c=1$, prove $a^2+b^2+c^2\ge1/3$ and $1/a+1/b+1/c\ge9$.

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Solution

Cauchy-Schwarz with the all-ones vector gives

$$1=(a+b+c)^2\le3(a^2+b^2+c^2).$$

The positive-denominator form gives the reciprocal bound directly, since the denominator sum is one. Both equalities occur at $a=b=c=1/3$. The square-sum bound extends to real $a,b,c$ of sum one, but the reciprocal argument requires positivity.

ExerciseApply Cauchy-Schwarz twice

For real $a_i,b_i,c_i$ and $n\ge1$, prove

$$\begin{aligned} &\left(\sum_{i=1}^{n}a_i b_i c_i\right)^4\\ &\quad\le\left(\sum_{i=1}^{n}a_i^4\right) \left(\sum_{i=1}^{n}b_i^2\right)^2\\ &\qquad\cdot\left(\sum_{i=1}^{n}c_i^4\right). \end{aligned}$$

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Solution

Write

$$\begin{aligned} U&=\sum_{i=1}^{n}a_i b_i c_i,\\ V&=\sum_{i=1}^{n}a_i^2c_i^2,\\ B&=\sum_{i=1}^{n}b_i^2. \end{aligned}$$

First $U^2\le VB$. Squaring preserves order because both sides are nonnegative. A second application gives

$$V^2\le\left(\sum_{i=1}^{n}a_i^4\right) \left(\sum_{i=1}^{n}c_i^4\right).$$

Combine the bounds. If $B=0$, equality is automatic. If $B>0$, equality requires both Cauchy-Schwarz steps to be equalities: $(b_i)$ and $(a_i c_i)$ must be linearly dependent, as must $(a_i^2)$ and $(c_i^2)$, with zero vectors included in that condition.

ExerciseA parameter bound and a missing maximum

For real $k$, show that

$$f_k(x)=\frac{(x+k)^2}{x^2+1}\le1+k^2$$

for every real $x$. When is the bound attained?

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Solution

Apply Cauchy-Schwarz to $(x,1)$ and $(1,k)$. Alternatively, the exact gap is

$$1+k^2-f_k(x)=\frac{(kx-1)^2}{x^2+1}\ge0.$$

If $k\ne0$, equality occurs at $x=1/k$. If $k=0$, every value is strictly below $1$, but $f_0(x)$ approaches $1$ as $|x|$ grows, so the supremum is $1$ and there is no maximum. No step divides by $x$, so the proof also covers $x=0$.

ExerciseA cyclic quadratic gap

For real $a,b,c,d$, prove $a^2+b^2+c^2+d^2\ge ab+bc+cd+da$ and find equality.

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Solution

Twice the difference between the two sides is

$$\begin{aligned} &(a-b)^2+(b-c)^2\\ &\quad+(c-d)^2+(d-a)^2. \end{aligned}$$

It is nonnegative, and it vanishes exactly when $a=b=c=d$. This directly identifies the equality condition without an additional estimate.

ExerciseA reciprocal sum with a necessary n-condition

Let $n\ge2$ and $a_i>0$, and define

$$s=\sum_{i=1}^{n}a_i.$$

Prove

$$\sum_{i=1}^{n}\frac{a_i}{s-a_i}\ge\frac{n}{n-1}.$$

What does this imply for three positive numbers, including triangle side lengths?

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Solution

All $s-a_i$ are positive, with sum $(n-1)s$. The reciprocal bound gives

$$\sum_{i=1}^{n}\frac1{s-a_i}\ge\frac{n^2}{(n-1)s}.$$

Consequently,

$$\begin{aligned} \sum_{i=1}^{n}\frac{a_i}{s-a_i} &=s\sum_{i=1}^{n}\frac1{s-a_i}-n\\ &\ge\frac{n^2}{n-1}-n =\frac n{n-1}. \end{aligned}$$

Equality holds exactly when all $a_i$ are equal. For $n=3$, the sum is at least $3/2$, hence strictly greater than $1$. Triangle inequalities are not needed; positivity alone suffices. At $n=1$, the original denominator is zero, explaining the restriction.

ExerciseChoose methods for three ratio inequalities

For positive $x,y,z$, prove

$$\frac{x^2}{y}+\frac{y^2}{x}\ge x+y,$$$$\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2} \ge\frac{x}{z}+\frac{y}{x}+\frac{z}{y},$$$$\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{zx}{y^2} \ge\frac{x}{y}+\frac{y}{z}+\frac{z}{x}.$$

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Solution

For the first, subtract the right side: the gap is $(x-y)^2(x+y)/(xy)\ge0$. Equality holds at $x=y$.

For the second, put $a=x/y$, $b=y/z$, $c=z/x$. It becomes $a^2+b^2+c^2\ge ab+bc+ca$, whose gap is half of $(a-b)^2+(b-c)^2+(c-a)^2$. Equality forces $a=b=c=1$, hence $x=y=z$.

For the third, multiply by positive $(xyz)^2$ and put $A=xy$, $B=yz$, $C=zx$. The desired statement becomes

$$\begin{aligned} A^3+B^3+C^3&\ge A^2B+B^2C\\ &\quad+C^2A. \end{aligned}$$

The lists of positive values $A,B,C$ and their squares have the same ordering. Rearrangement says the identity pairing is at least as large as this cyclic pairing. Equality requires $A=B=C$, hence $x=y=z$.

ExerciseExploration: logarithms turn products into ordered sums

For $a,b,c>0$, prove

$$a^a b^b c^c\ge a^b b^c c^a,$$$$a^a b^b c^c\ge(abc)^{(a+b+c)/3}.$$

Use that the natural logarithm is increasing and turns products into sums.

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Solution

Taking logarithms turns the first comparison into a same-order dot product versus a cyclic pairing of $a,b,c$ with $\ln a,\ln b,\ln c$. These two lists have the same ordering, so rearrangement proves it. One may sort the paired entries simultaneously; the cyclic assignment remains a permutation after sorting.

For the second, Chebyshev’s sum inequality gives

$$\begin{aligned} &a\ln a+b\ln b+c\ln c\\ &\quad\ge\frac{a+b+c}{3}(\ln a+\ln b+\ln c). \end{aligned}$$

Exponentiating preserves both comparisons. In either case equality holds exactly when $a=b=c$. Merely taking logarithms is a reformulation, so the rearrangement or Chebyshev step is needed to finish the proof.

ExerciseExploration: rational weights in AM-GM

For positive $a,b$, prove $a^{3/4}b^{1/4}\le(3a+b)/4$. Extend the argument to a weight $m/n$ with integers $0<m<n$.

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Solution

Apply four-term AM-GM to $a,a,a,b$. More generally, use $m$ copies of $a$ and $n-m$ copies of $b$:

$$a^{m/n}b^{1-m/n} \le\frac mn a+\left(1-\frac mn\right)b.$$

Equality holds exactly when $a=b$. This proves the rational-weight case directly; extending to arbitrary real weights requires an additional limiting or convexity argument.

ExerciseExploration: why a norm needs the triangle inequality

For $p>0$, define

$$N_p(u)=\left(\sum_{i=1}^{n}|u_i|^p\right)^{1/p}.$$

We proved the triangle inequality at $p=2$. Test it at $0<p<1$ using $u=(1,0)$ and $v=(0,1)$.

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Solution

Both $N_p(u)$ and $N_p(v)$ equal $1$, whereas $N_p(u+v)=2^{1/p}>2$ when $0<p<1$. Thus the triangle inequality fails, and this expression is not a norm in that range. The general Minkowski inequality is valid for $p\ge1$; its proof beyond the Euclidean case belongs with a fuller study of norms [2][2] P. S. Loh, “II. Inequalities,” 2003. June 18, 2003; notes on norms, rearrangement, and Chebyshev's sum inequality. https://www.math.cmu.edu/~ploh/docs/math/2-inequalities-solns.pdf.

Continue with Complex Numbers, where modulus gives a two-dimensional distance and the triangle inequality reappears geometrically.

References

1. [1] P. S. Loh, “6. Inequalities,” 2023. Carnegie Mellon University, Putnam Seminar, Fall 2023; classical inequalities. https://www.math.cmu.edu/~ploh/docs/math/2023-295/06-inequalities.pdf ↩
2. [2] P. S. Loh, “II. Inequalities,” 2003. June 18, 2003; notes on norms, rearrangement, and Chebyshev's sum inequality. https://www.math.cmu.edu/~ploh/docs/math/2-inequalities-solns.pdf a b