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Toolkit 28

Max-QM-AM-GM-HM-Min Inequality

x1,…,xn∈R+x_1,\dots,x_n\in\mathbb{R}^+
max⁡{xi}≥x12+x22+⋯+xn2n≥x1+x2+⋯+xnn≥x1x2⋯xnn≥n1x1+1x2+⋯+1xn≥min⁡{xi}\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}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}}\ge\min\{x_i\}
Equality holds if and only if all xi are equal.\text{Equality holds if and only if all }x_i\text{ are equal.}

Proof

Without loss of generality, assume that

x1≤x2≤⋯≤xn.x_1\le x_2\le\cdots\le x_n.

Thus,

min⁡{x1,x2,…,xn}=x1\min\{x_1,x_2,\ldots,x_n\}=x_1

and

max⁡{x1,x2,…,xn}=xn.\max\{x_1,x_2,\ldots,x_n\}=x_n.

1. Max–QM

We want to prove that

xn≥x12+x22+⋯+xn2n.x_n \ge \sqrt{\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}}.

Since

x1≤xn,x2≤xn,…,xn−1≤xn,x_1\le x_n,\quad x_2\le x_n,\quad \ldots,\quad x_{n-1}\le x_n,

we have

x12≤xn2,x22≤xn2,…,xn−12≤xn2.x_1^2\le x_n^2,\quad x_2^2\le x_n^2,\quad \ldots,\quad x_{n-1}^2\le x_n^2.

Together with xn2=xn2x_n^2=x_n^2, this gives

x12+x22+⋯+xn2≤nxn2.x_1^2+x_2^2+\cdots+x_n^2 \le nx_n^2.

Dividing by nn and taking square roots,

xn≥x12+x22+⋯+xn2n.x_n \ge \sqrt{\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}}.

2. QM–AM

We want to prove that

x12+x22+⋯+xn2n≥x1+x2+⋯+xnn.\sqrt{\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}} \ge \frac{x_1+x_2+\cdots+x_n}{n}.

Since

(x1−x2)2+(x1−x3)2+⋯+(xn−1−xn)2≥0,(x_1-x_2)^2+(x_1-x_3)^2+\cdots+(x_{n-1}-x_n)^2 \ge 0,

expanding all the squares gives

(n−1)(x12+x22+⋯+xn2)≥2x1x2+2x1x3+⋯+2xn−1xn.(n-1)(x_1^2+x_2^2+\cdots+x_n^2) \ge 2x_1x_2+2x_1x_3+\cdots+2x_{n-1}x_n.

Adding

x12+x22+⋯+xn2x_1^2+x_2^2+\cdots+x_n^2

to both sides, we obtain

n(x12+x22+⋯+xn2)≥(x1+x2+⋯+xn)2.n(x_1^2+x_2^2+\cdots+x_n^2) \ge (x_1+x_2+\cdots+x_n)^2.

Dividing by n2n^2,

x12+x22+⋯+xn2n≥(x1+x2+⋯+xnn)2.\frac{x_1^2+x_2^2+\cdots+x_n^2}{n} \ge \left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^2.

Taking square roots gives

x12+x22+⋯+xn2n≥x1+x2+⋯+xnn.\sqrt{\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}} \ge \frac{x_1+x_2+\cdots+x_n}{n}.

3. AM–GM

We prove

x1+x2+⋯+xnn≥x1x2⋯xnn\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=2n=2

We have

(x1−x2)2≥0.(\sqrt{x_1}-\sqrt{x_2})^2\ge 0.

Therefore,

x1+x2≥2x1x2,x_1+x_2\ge 2\sqrt{x_1x_2},

and hence

x1+x22≥x1x2.\frac{x_1+x_2}{2}\ge\sqrt{x_1x_2}.

Thus, AM–GM holds for n=2n=2.

Forward step: from nn to 2n2n

Assume AM–GM holds for nn positive real numbers.

Consider 2n2n positive real numbers

x1,x2,…,xn,xn+1,…,x2n.x_1,x_2,\ldots,x_n,x_{n+1},\ldots,x_{2n}.

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

x1+x2+⋯+xnn≥x1x2⋯xnn\frac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}

and

xn+1+xn+2+⋯+x2nn≥xn+1xn+2⋯x2nn.\frac{x_{n+1}+x_{n+2}+\cdots+x_{2n}}{n} \ge \sqrt[n]{x_{n+1}x_{n+2}\cdots x_{2n}}.

Therefore,

x1+x2+⋯+x2n2n=x1+⋯+xnn+xn+1+⋯+x2nn2≥x1x2⋯xnn+xn+1xn+2⋯x2nn2.\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} \\ &\ge \frac{\sqrt[n]{x_1x_2\cdots x_n}+\sqrt[n]{x_{n+1}x_{n+2}\cdots x_{2n}}}{2}. \end{aligned}

Applying the two-variable AM–GM inequality,

x1+x2+⋯+x2n2n≥x1x2⋯xnn xn+1xn+2⋯x2nn=x1x2⋯x2n2n.\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}}} \\ &= \sqrt[2n]{x_1x_2\cdots x_{2n}}. \end{aligned}

Thus, if AM–GM holds for nn, it also holds for 2n2n.

Backward step: from nn to n−1n-1

Assume AM–GM holds for nn positive real numbers.

Let x1,x2,…,xn−1x_1,x_2,\ldots,x_{n-1} be positive, and define

xn=x1+x2+⋯+xn−1n−1.x_n=\frac{x_1+x_2+\cdots+x_{n-1}}{n-1}.

Then

x1+x2+⋯+xn−1+xnn=xn.\frac{x_1+x_2+\cdots+x_{n-1}+x_n}{n} = x_n.

Applying AM–GM to the nn numbers x1,x2,…,xnx_1,x_2,\ldots,x_n, we get

xn≥x1x2⋯xn−1xnn.x_n \ge \sqrt[n]{x_1x_2\cdots x_{n-1}x_n}.

Raising both sides to the nn-th power,

xnn≥x1x2⋯xn−1xn.x_n^n \ge x_1x_2\cdots x_{n-1}x_n.

Since xn>0x_n>0, divide by xnx_n:

xnn−1≥x1x2⋯xn−1.x_n^{n-1} \ge x_1x_2\cdots x_{n-1}.

Taking the (n−1)(n-1)-st root gives

xn≥x1x2⋯xn−1n−1.x_n \ge \sqrt[n-1]{x_1x_2\cdots x_{n-1}}.

Substituting the definition of xnx_n,

x1+x2+⋯+xn−1n−1≥x1x2⋯xn−1n−1.\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 nn, it also holds for n−1n-1.

Starting from n=2n=2, the forward step proves the inequality for every power of 22, and the backward step then proves it for every positive integer nn.

Therefore,

x1+x2+⋯+xnn≥x1x2⋯xnn.\frac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}.

4. GM–HM

Apply AM–GM to the positive numbers

1x1,1x2,…,1xn.\frac{1}{x_1},\frac{1}{x_2},\ldots,\frac{1}{x_n}.

We obtain

1x1+1x2+⋯+1xnn≥1x1x2⋯xnn.\frac{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}}{n} \ge \sqrt[n]{\frac{1}{x_1x_2\cdots x_n}}.

Since both sides are positive, taking reciprocals reverses the inequality:

n1x1+1x2+⋯+1xn≤x1x2⋯xnn.\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}} \le \sqrt[n]{x_1x_2\cdots x_n}.

Therefore,

x1x2⋯xnn≥n1x1+1x2+⋯+1xn.\sqrt[n]{x_1x_2\cdots x_n} \ge \frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}}.

5. HM–Min

Since

x1≤x2≤⋯≤xn,x_1\le x_2\le\cdots\le x_n,

taking reciprocals gives

1x1≥1x2≥⋯≥1xn.\frac{1}{x_1}\ge\frac{1}{x_2}\ge\cdots\ge\frac{1}{x_n}.

In particular,

1x1≥1xi\frac{1}{x_1}\ge\frac{1}{x_i}

for every ii. Therefore,

1x1+1x2+⋯+1xn≤nx1.\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n} \le \frac{n}{x_1}.

Since both sides are positive, taking reciprocals reverses the inequality:

11x1+1x2+⋯+1xn≥x1n.\frac{1}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}} \ge \frac{x_1}{n}.

Multiplying by nn, we obtain

n1x1+1x2+⋯+1xn≥x1.\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}} \ge x_1.

Since x1=min⁡{x1,x2,…,xn}x_1=\min\{x_1,x_2,\ldots,x_n\},

n1x1+1x2+⋯+1xn≥min⁡{x1,x2,…,xn}.\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}} \ge \min\{x_1,x_2,\ldots,x_n\}.

Equality in all five inequalities holds if and only if

x1=x2=⋯=xn.□x_1=x_2=\cdots=x_n. \quad\square