AM–GM: proofs, equality conditions, and five worked problems

The inequality and its equality condition

For nonnegative real numbers $a,b$,

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

with equality exactly when $a=b$. The arithmetic mean is the average of the numbers; the geometric mean is the square root of their product. A short proof is

$$\frac{a+b}{2}-\sqrt{ab}=\frac12(\sqrt a-\sqrt b)^2\ge0.$$

The square vanishes exactly when $a=b$.

A geometric picture

Take a circle of diameter $a+b$, and divide that diameter into lengths $a$ and $b$. The perpendicular half-chord at the division point has length $\sqrt{ab}$ by the intersecting-chords theorem. It cannot exceed the radius $(a+b)/2$; equality occurs when the division point is the center, or $a=b$.

The degenerate cases $a=0$ or $b=0$ follow directly from the algebraic inequality.

The general form

For $n$ nonnegative numbers,

$$\boxed{\frac{a_1+\cdots+a_n}{n}\ge(a_1\cdots a_n)^{1/n},}$$

with equality exactly when all numbers are equal.

Here is a proof using $e^u\ge1+u$, whose equality condition is $u=0$. This auxiliary inequality follows because $e^u-1-u$ decreases before zero, increases afterward, and is zero at zero.

Let $M$ be the arithmetic mean. If $M=0$, all the numbers are zero and the conclusion is immediate. If $M>0$, apply the auxiliary inequality with $u=a_k/M-1$ and multiply the resulting nonnegative inequalities:

$$\prod_{k=1}^n\frac{a_k}{M} \le\exp\left(\sum_{k=1}^n\frac{a_k}{M}-n\right)=1.$$

Thus $a_1\cdots a_n\le M^n$. If any $a_k=0$ while $M>0$, the AM–GM inequality is strict. Otherwise equality in the product requires equality in every exponential inequality, hence $a_k=M$ for all $k$.

Three checks when using AM–GM for an extremum

Check that the terms are nonnegative, obtain a bound independent of the free variables, and verify that equality is attainable under all constraints.

For instance, when $x>0$,

$$x+\frac1x\ge2,$$

with equality at $x=1$, so the minimum is two. If the domain were $x\ge2$, that equality point would be unavailable; the minimum would instead be $5/2$ at $x=2$.

A moving lower bound is not a minimum

For $x>0$, AM–GM correctly gives

$$x^2+\frac1x\ge2\sqrt x,$$

with equality at $x=1$. It does not prove that the minimum is two, because the right side changes with $x$.

Instead, split the reciprocal:

$$x^2+\frac1x=x^2+\frac1{2x}+\frac1{2x} \ge3\sqrt[3]{\frac14}=\frac3{\sqrt[3]4}.$$

Equality requires $x^2=1/(2x)$, giving $x=1/\sqrt[3]2>0$. This proves the stated constant is the minimum.

Problem 1: divide first, then complete a useful pair

For $x>-1$, minimize $(x^2+x+4)/(x+1)$.

Polynomial division gives

$$\frac{x^2+x+4}{x+1}=x+\frac4{x+1} =(x+1)+\frac4{x+1}-1.$$

Both terms in the pair are positive, with product four. Therefore

$$(x+1)+\frac4{x+1}-1\ge4-1=3.$$

Equality requires $(x+1)^2=4$ and $x+1>0$, so $x=1$. Hence the minimum is three, attained at one. Adding and subtracting one created a pair with a constant product.

Problem 2: maximize by bounding a reciprocal

For $x>0$, maximize $x/(x^2-x+4)$.

The denominator is $(x-1/2)^2+15/4>0$, so the fraction is positive. Its reciprocal satisfies

$$\frac{x^2-x+4}{x}=x+\frac4x-1\ge4-1=3.$$

Equality occurs at $x=4/x$, or $x=2$. Taking reciprocals reverses the inequality between these positive numbers, giving

$$\boxed{\max\frac{x}{x^2-x+4}=\frac13\quad(x=2).}$$

The constant in the completed square is $15/4$; the source's different constant is a typographical error.

Problem 3: exploit homogeneity

For $x,y>0$, maximize

$$\frac{x^2+xy+y^2}{x^2-xy+y^2}.$$

Divide numerator and denominator by $y^2$, and set $t=x/y>0$. Every positive $t$ is attainable. Then

$$\frac{t^2+t+1}{t^2-t+1} =1+\frac2{t+1/t-1}\le1+\frac2{2-1}=3.$$

The denominator is positive. Equality holds exactly when $t=1$, or $x=y$. Thus the maximum is three, attained for every positive equal pair. Homogeneity reduced two free variables to their ratio.

Problem 4: a triangle with a moving point

Let $O=(0,0)$, $A=(0,1)$, $B=(1,0)$, and $C=(t,0)$, where $0<t<1$. Choose $D$ on segment $AB$ so that $\angle ACO=\angle BCD$. Find the maximum area of triangle $ACD$.

The slope of $CA$ is $-1/t$, so the equal-angle condition gives slope $1/t$ for $CD$. Intersect

$$y=1-x,\qquad y=\frac{x-t}{t}$$

to obtain

$$D=\left(\frac{2t}{1+t},\frac{1-t}{1+t}\right).$$

These coordinates do lie on $AB$ for $0<t<1$. The determinant area formula gives

$$\begin{aligned} [ACD]&=\frac{t(1-t)}{1+t} =3-\left(t+1+\frac2{t+1}\right)\\ &\le3-2\sqrt2. \end{aligned}$$

Equality requires $(t+1)^2=2$, so $t=\sqrt2-1$, which lies in the allowed interval. Therefore

$$\boxed{\max[ACD]=3-2\sqrt2.}$$

Problem 5: minimize a tetrahedron volume

Let $OABC$ be a regular tetrahedron of side one, and let $G$ satisfy

$$4\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}.$$

Points $P,Q,R$ lie in the interiors of $OA,OB,OC$, respectively, with

$$\overrightarrow{OP}=p\overrightarrow{OA},\qquad \overrightarrow{OQ}=q\overrightarrow{OB},\qquad \overrightarrow{OR}=r\overrightarrow{OC},\qquad0<p,q,r<1.$$

Suppose $G$ lies in triangle $PQR$. Prove $1/p+1/q+1/r=4$, find the perpendicular foot $H$ from $R$ to plane $OAB$, and minimize the volume of $OPQR$.

The plane condition

The vectors along the three edges at $O$ are linearly independent. Expressing $G$ using $OP,OQ,OR$ gives

$$\overrightarrow{OG}=\frac1{4p}\overrightarrow{OP} +\frac1{4q}\overrightarrow{OQ} +\frac1{4r}\overrightarrow{OR}.$$

Membership in the affine plane $PQR$ means these coefficients sum to one. Thus

$$\boxed{\frac1p+\frac1q+\frac1r=4.}$$

The coefficients are positive, so when the equation holds, $G$ is indeed inside the triangle, not merely on its extended plane.

The perpendicular foot

Write $OH=s\,OA+t\,OB$ in vector notation. Each edge vector has length one, and their pairwise inner products are $1/2$. The conditions $RH\cdot OA=RH\cdot OB=0$ become

$$s+\frac t2-\frac r2=0,\qquad \frac s2+t-\frac r2=0.$$

Hence $s=t=r/3$, and

$$\boxed{\overrightarrow{OH}=\frac r3(\overrightarrow{OA}+\overrightarrow{OB}).}$$

Notice the minus signs on the $r$ terms; they are required because $RH=OH-OR$. Computing the squared length yields $|RH|^2=2r^2/3$.

The volume bound

The base $OPQ$ has area $pq\sqrt3/4$, and the height is $r\sqrt{2/3}$. Therefore

$$V=\frac{\sqrt2}{12}pqr.$$

AM–GM applied to the positive reciprocals gives

$$4=\frac1p+\frac1q+\frac1r\ge3\sqrt[3]{\frac1{pqr}}, \qquad pqr\ge\left(\frac34\right)^3.$$

Equality is attained at $p=q=r=3/4$, which satisfies the segment constraints and the triangle condition. Thus

$$\boxed{V_{\min}=\frac{9\sqrt2}{256}.}$$

In each problem, the bound alone is only part of the solution. Its equality condition is what turns an inequality into an attained minimum or maximum.