# Quod Erat Demonstrandum

## 2017/08/13

### 最小值

Filed under: Additional / Applied Mathematics,NSS,Pure Mathematics — johnmayhk @ 6:19 下午

$x\ge 3$

$x$ 的最小值是 3，

$f(x)$ 的最小值不是 3，而是 4。

$x+\frac{1}{x} \ge 2$

$x+\frac{1}{x}=(\sqrt{x}-\sqrt{\frac{1}{x}})^2+2$

$x+\frac{1}{x} \ge 2$

$a+\frac{1}{a} \ge 2\Rightarrow (a+\frac{1}{a})^2 \ge 4$

$(b+\frac{1}{b})^2 \ge 4$

$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2\ge 8$ ……………….. (*)

$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ 的最小值是 8 嗎？錯！

https://www.wolframalpha.com/input/?i=minimum+(a%2B1%2Fa)%5E2%2B(b%2B1%2Fb)%5E2+for+a%2Bb%3D1+and+a%3E0+and+b%3E0

$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ 不能等於 8。

$(a+\frac{1}{a})^2 = (b+\frac{1}{b})^2 = 4$

$a=b=1$

$a+b=1$

$PO \ge PQ$

https://www.geogebra.org/m/KkcCjPGT

$\sqrt{(a+\frac{1}{a})^2+(b+\frac{1}{b})^2}\ge \frac{|(a+\frac{1}{a})+(b+\frac{1}{b})|}{\sqrt{1^2+1^2}}$

$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2\ge \frac{(a+\frac{1}{a}+b+\frac{1}{b})^2}{2}=\frac{1}{2}(1+\frac{1}{ab})^2$（因為 $a+b=1$

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

$(\sqrt{a}-\sqrt{b})^2 \ge 0\Rightarrow a+b-2\sqrt{ab}\ge 0$

$1\ge 2\sqrt{ab}\Rightarrow \frac{1}{ab}\ge 4$

$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2\ge \frac{1}{2}(1+4)^2 = \frac{25}{2}$

$a=b=\frac{1}{2}$

$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2=(\frac{5}{2})^2+(\frac{5}{2})^2=\frac{25}{2}$，數值可達，

$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ 的最小值是 $\frac{25}{2}$

$\displaystyle \sum_{i=1}^n (a_i+\frac{1}{a_i})^m\ge n(\frac{k}{n}+\frac{n}{k})^m$

$f(x)=(x+\frac{1}{x})^2+(1-x+\frac{1}{1-x})^2$

https://www.wolframalpha.com/input/?i=derivative+(x%2B1%2Fx)%5E2%2B(1-x%2B1%2F(1-x))%5E2

$x=\frac{1}{2}$

$x=\frac{1}{6}(2-2^{\frac{2}{3}}\sqrt[3]{25-3\sqrt{69}}-2^{\frac{2}{3}}\sqrt[3]{25+3\sqrt{69}}) < 0$

$x=\frac{1}{6}(4+2^{\frac{2}{3}}\sqrt[3]{25-3\sqrt{69}}+2^{\frac{2}{3}}\sqrt[3]{25+3\sqrt{69}}) > 1$

$f''(\frac{1}{2})=196 > 0$

wolframalpha 告之