这里将会持续添加一些基础的不等式定理、题目。

Basic Inequalities

Theorem1. 1

xy2xyxy\geq2\sqrt{xy}. x,yR+x,y\in \R ^+

Proof

(xy)20x22xy+y20x2+2xy+y24xy(x+y)24xyx+y2xyxyx+y2Equalitiy  occurs  if  and  only  if  x=y.\begin{aligned} (x-y)^2&\geq0\\ x^2-2xy+y^2&\geq0\\ x^2+2xy+y^2&\geq4xy\\ (x+y)^2&\geq4xy\\ x+y&\geq2\sqrt{xy}\\ \Leftrightarrow xy&\leq\frac{x+y}{2}\\ Equalitiy\ \ occurs\ \ if\ \ and\ \ only\ \ if\ \ x=y. \end{aligned}

Exercise1. 1

Use Theorem1. 1 to solve this problem.

Let 0<x<40<x<4 . Prove the inequality:

x(82x)8x(8-2x)\leq8 .

Solution

2x+(82x)=8x(82x)=12[2x(82x)]12(2x82x2)2=8\begin{aligned} 2x+(8-2x)=8\\ x(8-2x)=\frac{1}{2}[2x(8-2x)]\leq\frac{1}{2}(\frac{2x-8-2x}{2})^2=8 \end{aligned}

Exercise1.2

Let nRn\in R . Prove the inequality:

1+122+132+...+1n221+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\leq2 .

Solution

It’s easy to know:

1a×a1a×(a1).\frac{1}{a\times a}\leq\frac{1}{a\times(a-1)}.

So

1+122+132+...+1n2<1+11×2+12×3+...+1(n1)×n=1+1112+1213+...+1n11n=21n<2\begin{aligned} 1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}<1+\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{(n-1)\times n}\\ =1+\frac1 1 -\frac1 2 +\frac1 2-\frac1 3+...+\frac{1}{n-1}-\frac{1}{n}=2-\frac{1}{n}<2 \end{aligned}

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