2027 Strong Foundation Mathematics: AM-GM Inequalities

Example 2.1

It is known that $a > b > c$makes the inequality

$$\frac{1}{a-b} + \frac{1}{b-c} \geq \frac{k}{a-c}$$

The maximum value of a constant real number $k$is ______.

The denominator is complicated, so change the denominator:

$$\begin{gathered} a-b=x,b-c=y,a-c=x+y\\ \frac{1}{x}+\frac{1}{y}\ge\frac{(1+1)^2}{x+y}\Longrightarrow k\le4 \end{gathered}$$

Example 2.2

Assume $a, b, c$is a positive real number, prove:

$$\frac{a + 3c}{a + 2b + c} + \frac{4b}{a + b + 2c} - \frac{8c}{a + b + 3c} \geq 12\sqrt{2} - 17.$$

The denominator is complicated, so change the denominator:

$$\begin{gathered} a+2b+c=x,a+b+2c=y,a+b+3c=z\\ \begin{cases} a=-x+5y-3z,\\ b=x-2y+z,\\ c=0x-y+z \end{cases}\\ \frac{-x+2y}{x}+\frac{4x-8y+4z}{y}+\frac{8y-8z}{z}\\ =-17+2\frac{y}{x}+4\frac{x}{y}+4\frac{z}{y}+8\frac{y}{z}\\ \ge-17+2\sqrt{2\cdot4}+2\sqrt{4\cdot8}=-17+12\sqrt{2} \end{gathered}$$

Example 2.3

Given $a > b > 0$, then the minimum value of $a^2 + \frac{1}{b(a-b)}$is

A. 2

B. 4

C. $2\sqrt{5}$

D. 5

Consider b first: $$\begin{gathered} a^2 + \frac{1}{b(a-b)}\\ \ge a^2+\frac{4}{a^2}\ge4 \end{gathered}$$

Or consider incremental substitution:

$$\begin{gathered} b=x,a=x+y\\ (x+y)^2+\frac{1}{xy}\\ \ge4xy+\frac{1}{xy}\ge4 \end{gathered}$$

Example 2.4

(2012 Tsinghua Summer Camp) It is known that $a, b, c$is the side length of the three sides of the triangle $\triangle ABC$, then the following judgment is correct:

A. $\frac{3}{2} \leq \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} < 2$

B. $\frac{3}{2} \lt \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} < 2$

C. $\frac{3}{2} \lt \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \leq 2$

D. $\frac{3}{2} \leq \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \leq 2$

$$\begin{gathered} a\lt b+c\Longrightarrow \frac{2a}{a+b+c}\gt\frac{a}{b+c}\\ \end{gathered}$$

Rotate $a \to b \to c \to a$and get:

$$b < c+a \quad\Longrightarrow\quad \frac{2b}{a+b+c} > \frac{b}{c+a}$$

$$c < a+b \quad\Longrightarrow\quad \frac{2c}{a+b+c} > \frac{c}{a+b}$$

Accumulation: $S\lt2$

Let's consider the lower bound, repeat the same trick, change the denominator, and use the AM-GM inequality:

$$\begin{gathered} b+c=x,c+a=y,a+b=z\\ S=\frac{\frac{y+z-x}{2}}{x}+\frac{\frac{z+x-y}{2}}{y}+\frac{\frac{x+y-z}{2}}{z}\\ =\frac{1}{2}(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z})-\frac{3}{2}\\ \ge 3-\frac{3}{2}=\frac{3}{2} \end{gathered}$$

Choose A.

Example 2.5

If the positive number $x, y$satisfies $x^2 + 2xy - 1 = 0$, then the minimum value of $2x + y$is

A. $\frac{\sqrt{2}}{2}$

B. $\sqrt{2}$

C. $\frac{\sqrt{3}}{2}$

D. $\sqrt{3}$

$$\begin{gathered} 2x+y=u,y=u-2x\\ x^2+2(u-2x)x-1=0\\ 3x^2-2ux+1=0\\ \Delta=4u^2-12\ge0\\ u\ge\sqrt{3}(x=y=\frac{\sqrt{3}}{3}) \end{gathered}$$

Or consider exchanging yuan: $$\begin{gathered} x(x+2y)=1\\ 2x+y=\frac{3}{2}x+\frac{1}{2}(x+2y)\ge\sqrt{3x(x+2y)}=\sqrt{3} \end{gathered}$$

Example 2.6

If the positive number $a, b, c$satisfies $a(a+b+c)+bc=4$, then the minimum value of the formula $3a+2b+c$is ______.

$$\begin{gathered} a^2+ab+ac+bc=(a+b)(a+c)=4\\ 3a+2b+c=2(a+b)+(a+c)\\\ge2\sqrt{2(a+b)(a+c)}=4\sqrt{3} \end{gathered}$$

Example 2.7

Assume $x, y, z > 0$, then

$$\frac{xy + 2yz}{x^2 + y^2 + z^2}$$

The maximum value is ______.

$$\begin{gathered} \frac{xy + 2yz}{x^2 + y^2 + z^2}\\ =\frac{xy + 2yz}{x^2 + \frac{1}{5}y^2+\frac{4}{5}y^2 + z^2}\\ \le \frac{xy+2yz}{2\sqrt{\frac{1}{5}}xy+2\sqrt{\frac{4}{5}yz}}\\ \le\frac{\sqrt{5}}{2} \end{gathered}$$

Example 2.8

Given $a \geq 0, b \geq 0, c \geq 0, a + b + c = 1$, find the maximum value of $\sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1}$.

Round up the mean by averaging:

$$\begin{gathered} \sqrt{(4a+1)\frac{7}{3}}\le\frac{4a+1+\frac{7}{3}}{2} \end{gathered}$$Simplification factor: $$\begin{gathered} \sqrt{4a+1}\le\frac{2\sqrt{21}}{7}a+\frac{5\sqrt{21}}{7}\\ \sqrt{4b+1}\le\frac{2\sqrt{21}}{7}b+\frac{5\sqrt{21}}{7}\\ \sqrt{4c+1}\le\frac{2\sqrt{21}}{7}c+\frac{5\sqrt{21}}{7}\\ S\le\frac{2\sqrt{21}}{7}(a+b+c)+\frac{5\sqrt{21}}{7}=\sqrt{21} \end{gathered}$$

Example 2.9

(Zhejiang University) Suppose the sum of positive numbers $x_1, x_2, \dots, x_n$is equal to 1 ($n \geq 2$), prove:

$$\frac{1}{x_1 - x_1^3} + \frac{1}{x_2 - x_2^3} + \cdots + \frac{1}{x_n - x_n^3} > 4$$

Consider local inequalities:

$$\begin{gathered} \frac{1}{x-x^3}\ge4x,x\in(0,1)\\ \Longleftrightarrow (2x^2-1)^2\ge0(x=\frac{\sqrt{2}}{2}) \end{gathered}$$

Obviously, it is impossible for every $x$to be $\frac{\sqrt{2}}{2}$, Q.E.D.

Example 2.10

Given $a_i > 0$, and $a_1a_2a_3\cdots a_n = 1$, find the minimum value of $(a_1+2)(a_2+2)\cdots(a_n+2)$.

Still holding the hand holding the equal sign: $$\begin{gathered} (a_1+1+1)(a_2+1+1)\cdots(a_n+1+1)\\ \ge3^n\sqrt[3]{a_1a_2a_3\cdots a_n}=3^n \end{gathered}$$

Example 2.11

Given $x > 0$, find the minimum value of $x^2 + \frac{16}{x}$.

$$\begin{gathered} x^2+\frac{8}{x}+\frac{8}{x}\\ \ge3\sqrt[3]{8^2}=12(x=2) \end{gathered}$$

Example 2.12

Given $x\gt0$, find the minimum value of $x^2+\frac{1}{x^3}$.

$$\begin{gathered} \frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{1}{2x^3}+\frac{1}{2x^3}\\ \ge5\sqrt[5]{\frac{1}{3^3\cdot2^2}}=\frac{5}{6}\sqrt[5]{72} \end{gathered}$$

Example 2.13

Given $0\lt x\lt1$, find the maximum value of $x^2\sqrt{1-x}$.

The lower bound is obviously 0, there is no minimum value, consider the maximum value:

$$\begin{gathered} x^2\sqrt{1-x}\\ =\sqrt{x^4(1-x)}\\ =\frac{1}{2}\sqrt{x\cdot x\cdot x\cdot x(4-4x)}\le\frac{1}{2}\sqrt{(\frac{4}{5})^5}=\frac{16\sqrt{5}}{125}(x=\frac{4}{5}) \end{gathered}$$

Example 2.14

Assume $x > -\frac{1}{2}$, then the minimum value of $f(x) = x^2 + x + \frac{4}{2x+1}$is ______.

Change denominator: $t=2x+1\gt0$

$$\begin{gathered} f(x)=(\frac{t-1}{2})^2+\frac{t-1}{2}+\frac{4}{t}\\ =\frac{t^2}{4}+\frac{4}{t}-\frac{1}{4}\\ =\frac{t^2}{4}+\frac{2}{t}+\frac{2}{t}-\frac{1}{4}\ge3-\frac{1}{4}=\frac{11}{4}(t=2,x=\frac{1}{2}) \end{gathered}$$

Example 2.15

Assume $a + 3b = 3$, and find the minimum value of the algebraic expression $3^a + 9^b$.

$$\begin{gathered} 3^a+9^b=3^a+3^{2b}\\ =\frac{3^a}{2}+\frac{3^a}{2}+\frac{3^{2b}}{3}+\frac{3^{2b}}{3}+\frac{3^{2b}}{3}\\ \ge5\sqrt[5]{\frac{3^{2a+6b}}{2^2\cdot3^3}}=5\sqrt[5]{\frac{3^3}{2^2}}=\frac{5}{2}(15)^\frac{3}{5} \end{gathered}$$

Example 2.16

Assume $a > b > 0$, and prove $\sqrt{2a^3} + \frac{3}{ab - b^2} \ge 10$.

① $\ge \sqrt{2} \cdot (2\sqrt{xy})^3 + \frac{3}{xy}$$$= 8\sqrt{2}(xy)^{\frac{3}{2}} + \frac{3}{xy}$$$$= 4\sqrt{2}(xy)^{\frac{3}{2}} + 4\sqrt{2}(xy)^{\frac{3}{2}} + \frac{3}{xy}$$$$\ge 5\sqrt[5]{32} = 10$$② $\ge \sqrt{2}(x+y)^3 + \frac{12}{(x+y)^2}$$$= \frac{\sqrt{2}}{2}(x+y)^3 + \frac{\sqrt{2}}{2}(x+y)^3 + \frac{4}{(x+y)^2} + \frac{4}{(x+y)^2} + \frac{4}{(x+y)^2}$$$$\ge 5\sqrt[5]{32} = 10$$

Example 2.17

Assume $a,b,c$is a positive real number, and $a + b + c = 1$, prove: $\frac{1}{a+bc} + \frac{1}{b+ac} + \frac{1}{c+ab} \ge \frac{27}{4}$.

The equal sign is obviously $a=b=c=\frac{1}{3}$. $$\begin{gathered} \frac{1}{a+bc}\\ =\frac{1}{a(a+b+c)+bc}\\ =\frac{1}{(a+b)(a+c)}\\ S\ge3\sqrt[3]{\frac{1^3}{(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)}}\\ \ge3\sqrt[3]{\frac{1}{(\frac{4}{6})^6}}=\frac{27}{4} \end{gathered}$$

Example 2.18

It is known that the positive real number $a,b$satisfies $ab(a+b) = 4$, then the minimum value of $2a + b$is $\underline{\quad\quad}$.

$$\begin{gathered} 4uv=(ua)(vb)(a+b)\le(\frac{(u+1)a+(v+1)b}{3})^3\\ \frac{u+1}{v+1}=2,ua=vb=a+b=k\\ u=2v+1,a=\frac{k}{2v+1},b=\frac{k}{v}\\ \frac{1}{v}+\frac{1}{2v+1}=1\\ \Longrightarrow v=\frac{1+\sqrt{3}}{2},u=2+\sqrt{3}\\ \frac{\frac{3+\sqrt{3}}{2}(2a+b)}{3}\ge \sqrt[3]{2(\sqrt{3}+1)(\sqrt{3}+2)}=1+\sqrt{3}\\ 2a+b\ge2\sqrt{3} \end{gathered}$$

Or a more attention-demanding solution:

$$\begin{gathered} (2a+b)^2=4a(a+b)+b^2\ge\frac{16}{b}+b^2\ge12\\ \Longrightarrow 2a+b\ge2\sqrt{3} \end{gathered}$$

Example 2.19

Assume $S_n = \sum_{k=1}^n \frac{1}{k}$, prove: $n(n+1)^{1/n} - n < S_n < n - (n-1)n^{-1/(n-1)} \quad (n > 1)$.

Key:1 pairing $$\begin{gathered} S_n+n=\frac{2}{1}+\frac{3}{2}+\frac{4}{3}+\cdots+\frac{n+1}{n}\ge n(n+1)^\frac{1}{n}\\ n-S_n=0+\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n}\ge(n-1)(\frac{1}{n})^\frac{1}{n-1} \end{gathered}$$

Example 2.20

Proof: $1 - \frac{1}{2012} \left( \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2013} \right) > \frac{1}{\sqrt[2012]{2013}}$.

Still note the pairing of 1: $$\begin{gathered} \Longleftrightarrow 2012-(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2013})\gt 2012\sqrt[2013]{\frac{1}{2023}}\\ \Longleftrightarrow \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots+\frac{2012}{2013}\gt 2012\sqrt[2013]{\frac{1}{2023}} \end{gathered}$$

Example 2.21

It is known that $m, n$is a positive integer, and $1 < m < n$, prove: $(1+m)^n > (1+n)^m$. $$\begin{gathered} (1+n)^m\cdot\underbrace{1 \times 1 \times \cdots \times 1}_{n-m\text{ terms}}\\ \lt(\frac{(n-m)+m(1+n)}{n})^n=(1+m)^n \end{gathered}$$

Example 2.22

Suppose $x_1, x_2, \dots, x_{n+1} > 0$satisfies $\frac{1}{1+x_1} + \frac{1}{1+x_2} + \dots + \frac{1}{1+x_{n+1}} = 1$, and verify: $x_1x_2\cdots x_{n+1} \ge n^{n+1}$.

Equality condition: all n.

Use the substitution method to use conditions: $$\begin{gathered} \frac{1}{1+x_i}=a_i\\ x_i=\frac{1}{a_i}-1\\ x_i=\frac{a_1+a_2+\cdots+a_{n+1}}{a_i}-1\\ x_i=\frac{a_1+a_2+\cdots+a_{i-1}+a_{i+1}+\cdots+a_{n+1}}{a_i}\\ \ge\frac{n\sqrt[n]{a_1a_2\cdots a_{i-1}a_{i+1}\cdots a_{n+1}}}{a_i}(i=1,2,3,\cdots,n+1) \end{gathered}$$

The cumulative multiplication of all is proved