Mean Inequalities

Lesson · Intermediate

Algebra: Inequalities

x_1,\ldots,x_n\in\mathbb{R}^+

\max\{x_i\} \ge \sqrt{\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}} \ge \frac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n} \ge \frac{n}{ \frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_n} } \ge \min\{x_i\}.

Equality holds if and only if all $x_i$ are equal.

Without loss of generality, assume that

x_1\le x_2\le\cdots\le x_n.

Thus,

\min\{x_1,x_2,\ldots,x_n\}=x_1

and

\max\{x_1,x_2,\ldots,x_n\}=x_n.

\frac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}.

We prove

\frac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}

by Cauchy's forward–backward induction.

Base case: $n=2$

We have

(\sqrt{x_1}-\sqrt{x_2})^2\ge0.

Therefore,

x_1+x_2\ge2\sqrt{x_1x_2},

and hence

\frac{x_1+x_2}{2} \ge \sqrt{x_1x_2}.

Thus, AM–GM holds for $n=2$ .

Forward step: from $n$ to $2n$

Assume AM–GM holds for $n$ positive real numbers.

Consider $2n$ positive real numbers

x_1,x_2,\ldots,x_n,x_{n+1},\ldots,x_{2n}.

Divide them into two groups of $n$ numbers. By the induction hypothesis,

\frac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}

and

\frac{x_{n+1}+x_{n+2}+\cdots+x_{2n}}{n} \ge \sqrt[n]{x_{n+1}x_{n+2}\cdots x_{2n}}.

Therefore,

\begin{aligned} \frac{x_1+x_2+\cdots+x_{2n}}{2n} &= \frac{ \frac{x_1+\cdots+x_n}{n} + \frac{x_{n+1}+\cdots+x_{2n}}{n} }{2} \\[4pt] &\ge \frac{ \sqrt[n]{x_1\cdots x_n} + \sqrt[n]{x_{n+1}\cdots x_{2n}} }{2}. \end{aligned}

Applying the two-variable AM–GM inequality,

\begin{aligned} \frac{x_1+x_2+\cdots+x_{2n}}{2n} &\ge \sqrt{ \sqrt[n]{x_1x_2\cdots x_n} \sqrt[n]{x_{n+1}x_{n+2}\cdots x_{2n}} } \\[4pt] &= \sqrt[2n]{x_1x_2\cdots x_{2n}}. \end{aligned}

Thus, if AM–GM holds for $n$ , it also holds for $2n$ .

Backward step: from $n$ to $n-1$

Assume AM–GM holds for $n$ positive real numbers.

Let $x_1,x_2,ldots,x_{n-1}$ be positive, and define

x_n = \frac{x_1+x_2+\cdots+x_{n-1}}{n-1}.

Then

\frac{x_1+x_2+\cdots+x_{n-1}+x_n}{n} = x_n.

Applying AM–GM to the $n$ numbers $x_1,x_2,ldots,x_n$ , we get

x_n \ge \sqrt[n]{x_1x_2\cdots x_{n-1}x_n}.

Raising both sides to the $n$ -th power,

x_n^n \ge x_1x_2\cdots x_{n-1}x_n.

Since $x_n>0$ , divide by $x_n$ :

x_n^{n-1} \ge x_1x_2\cdots x_{n-1}.

Taking the $(n-1)$ -st root gives

x_n \ge \sqrt[n-1]{x_1x_2\cdots x_{n-1}}.

Substituting the definition of $x_n$ ,

\frac{x_1+x_2+\cdots+x_{n-1}}{n-1} \ge \sqrt[n-1]{x_1x_2\cdots x_{n-1}}.

Thus, if AM–GM holds for $n$ , it also holds for $n-1$ .

Starting from $n=2$ , the forward step proves the inequality for every power of $2$ , and the backward step then proves it for every positive integer $n$ .

Therefore,

\frac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}.

Given $a\ge3$ . Find the minimum value of

S=a+\frac1a.

Common mistake:

S = a+\frac1a \ge 2\sqrt{a\cdot\frac1a} = 2 \quad\Longrightarrow\quad \min S=2.

Cause:

\min S=2 \iff a=\frac1a=1,

which contradicts the assumption $a\ge3$ .

Analyzing and finding the solution

Consider the variation table of $a$ , $\frac1a$ , and $S$ to estimate $\min S$ :

$a$	3	4	5	6	7	8	9	10	11	12	$\cdots$	30
$\frac1a$	$\frac13$	$\frac14$	$\frac15$	$\frac16$	$\frac17$	$\frac18$	$\frac19$	$\frac1{10}$	$\frac1{11}$	$\frac1{12}$	$\cdots$	$\frac1{30}$
$S$	$3\frac13$	$4\frac14$	$5\frac15$	$6\frac16$	$7\frac17$	$8\frac18$	$9\frac19$	$10\frac1{10}$	$11\frac1{11}$	$12\frac1{12}$	$\cdots$	$30\frac1{30}$

As we can see from the Variation Table, if $a$ increases, $S$ increases, so $S$ receives the least value at $a=3$ . We say

\min S=\frac{10}{3}

occurs at the point of incidence $a=3$ .

Since in AM–GM inequality the equality occurs when all joining numbers are equal, at the point of incidence $a=3$ we cannot use AM–GM directly for $a$ and $\frac1a$ , because

3\ne\frac13.

We assume that AM–GM inequality is used for the couple

\left( \frac{a}{\alpha}, \frac1a \right)

such that at the point of incidence $a=3$ occurs

\frac{a}{\alpha} = \frac1a.

That means the following Point of Incidence Pattern holds:

a=3 \quad\Longrightarrow\quad \frac3\alpha=\frac13 \quad\Longrightarrow\quad \alpha=9.

Point of incidence coefficient.

We transform $S$ according to the Point of Incidence Pattern as above.

Right solution

\begin{aligned} S &= a+\frac1a \\ &= \left( \frac{a}{9}+\frac1a \right) + \frac{8a}{9} \\[4pt] &\ge 2\sqrt{ \frac{a}{9}\cdot\frac1a } + \frac{8a}{9} \\[4pt] &\ge \frac23+\frac{8\cdot3}{9} = \frac{10}{3}. \end{aligned}

For $a=3$ , then

\min S=\frac{10}{3}.

Given

a,b>0, \qquad a+b\le1.

Find the minimum value of the expression

S=ab+\frac1{ab}.

Common mistake:

S = ab+\frac1{ab} \ge 2\sqrt{ ab\cdot\frac1{ab} } = 2 \quad\Longrightarrow\quad \min S=2.

Cause:

\min S=2 \iff ab=\frac1{ab} \iff \sqrt{ab}=1 \Longrightarrow \frac{a+b}{2}\ge1 \Longrightarrow 1\le\frac12,

which is impossible.

Analyzing and finding the solution:

The expression $S$ contains two variables $a,b$ , but if we take

t=ab \qquad\text{or}\qquad t=\frac1{ab},

the expression

S=t+\frac1t

will contain only one variable. When changing the variables we must find the defined area of the new variable as follows:

t=\frac1{ab} \quad\Longrightarrow\quad ab=\frac1t

and

t = \frac1{ab} \ge \frac1{ \left(\frac{a+b}{2}\right)^2 } \ge \frac1{ \left(\frac12\right)^2 } = 4.

Problem will be: Given $t\ge4$ . Find the minimum value of the expression

S=t+\frac1t.

Point of Incidence Pattern:

t=4 \quad\Longrightarrow\quad \frac4\alpha=\frac14 \quad\Longrightarrow\quad \alpha=16.

Point of incidence coefficient.

General solution:

\begin{aligned} S &= t+\frac1t \\ &= \left( \frac{t}{16}+\frac1t \right) + \frac{15t}{16} \\[4pt] &\ge 2\sqrt{ \frac{t}{16}\cdot\frac1t } + \frac{15t}{16} \\[4pt] &\ge \frac12 + \frac{15\cdot4}{16} = \frac{17}{4}. \end{aligned}

For $t=4$ or $a=b=\frac12$ , then

\min S=\frac{17}{4}.

Reduced solution: Since $t=4\iff a=b=\frac12$ , we transform $S$ directly as follows:

\begin{aligned} S &= ab+\frac1{ab} \\ &= \left( ab+\frac1{16ab} \right) + \frac{15}{16ab} \\[4pt] &\ge 2\sqrt{ ab\cdot\frac1{16ab} } + \frac{15}{ 16\left(\frac{a+b}{2}\right)^2 } \\[4pt] &\ge \frac12+\frac{15}{4} = \frac{17}{4}. \end{aligned}

For $a=b=\frac12$ , then

\min S=\frac{17}{4}.

Suppose we want to prove

A\ge B.

If we can find a smaller quantity $C$ such that

A\ge C,

then it is enough to prove

C\ge B,

because

A\ge C\ge B \quad\Longrightarrow\quad A\ge B.

However, be careful not to weaken the inequality too much. If instead

A\ge B\ge C,

then

C\le B.

So trying to prove

C\ge B

will fail.

This does not mean that the original inequality $A\ge B$ is false. It only means that the intermediate bound $C$ is too weak to prove it.

Suppose $a,b,c>0$ and

a+b+c=3.

Prove

\frac1{a^2+b^2+c^2} \le \frac13.

By AM–GM inequality,

\frac{a^2+b^2}{2}\ge ab,

\frac{a^2+c^2}{2}\ge ac,

\frac{b^2+c^2}{2}\ge bc.

Adding them,

a^2+b^2+c^2 \ge ab+ac+bc \qquad\qquad\text{(1)}

\frac1{a^2+b^2+c^2} \le \frac1{ab+ac+bc}.

If we prove

\frac1{ab+ac+bc} \le \frac13,

the problem will be solved.

\frac1{ab+ac+bc} \le \frac13

\Longleftrightarrow \quad ab+ac+bc \ge 3 = \frac{(a+b+c)^2}{3}

\Longleftrightarrow \quad 3ab+3ac+3bc \ge a^2+b^2+c^2+2ab+2ac+2bc

\Longleftrightarrow \quad ab+ac+bc \ge a^2+b^2+c^2.

In (1) we proved the reverse of that, so it gives a contradiction.

Here we weakened the inequality. In fact,

\frac1{a^2+b^2+c^2} \le \frac13 \le \frac1{ab+ac+bc}.

Here is the correct proof.

By (1),

a^2+b^2+c^2 \ge ab+ac+bc.

Multiplying by $2$ ,

2a^2+2b^2+2c^2 \ge 2ab+2ac+2bc.

Adding $a^2+b^2+c^2$ ,

3a^2+3b^2+3c^2 \ge a^2+b^2+c^2+2ab+2ac+2bc = (a+b+c)^2 = 3^2 = 9.

Dividing by $3$ ,

a^2+b^2+c^2 \ge 3.

So,

\frac1{a^2+b^2+c^2} \le \frac13.