2 replies to this topic

[Royal Sky]

1.Cho $1\leq a,b,c\leq 2$

CMR: $\left ( a+b+c \right )\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\leq 10$

2.CMR : với mọi n $\epsilon Z+$ ta co:

   $\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{(n+1)\sqrt{n}}< 2$

Edited by HappyLife, 13-02-2016 - 22:09.

[royal1534]

1.Cho $1\leq a,b,c\leq 2$

CMR: $\left ( a+b+c \right )\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\leq 10$

 

 Không mất tính tổng quát giả sử $a \geq b \geq c $

$\rightarrow (a-b)(b-c) \geq 0$

$\rightarrow b^2+ac \leq ab+bc$

$\rightarrow \frac{b}{a}+\frac{c}{b} \leq 1+\frac{c}{a}$
Tương tự ta cũng có $\frac{a}{b}+\frac{b}{c} \leq 1+\frac{a}{c}$

$\rightarrow Q=3+(\frac{a}{b}+\frac{b}{c})+(\frac{b}{a}+\frac{c}{b})+\frac{a}{c}+\frac{c}{a} \leq 5+2(\frac{a}{c}+\frac{c}{a})$
Vậy ta chỉ cần chứng minh $\frac{a}{c}+\frac{c}{a} \leq \frac{5}{2}$

$\leftrightarrow (a-2c)(a-c) \leq 0$ (Đúng vì $1 \leq \frac{a}{c} \leq 2$)

Chứng minh hoàn tất.Đẳng thức xảy ra khi $(a,b,c)$ là hoán vị của bộ $(1,1,2)$ hoặc $(1,2,2)$

[Quoc Tuan Qbdh]

 

2.CMR : với mọi n $\epsilon Z+$ ta co:

   $\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{(n+1)\sqrt{n}}< 2$

Ta có :
$\frac{1}{(n+1)\sqrt{n}}=\frac{\sqrt{n}}{(n+1)n}=\sqrt{n}(\frac{1}{n}-\frac{1}{n+1})=\sqrt{n}(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}})$

$<\sqrt{n}(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}})=2(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})$          

Suy ra : $\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{(n+1)\sqrt{n}}<2(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+.....-\frac{1}{\sqrt{n+1}})=2.(1-\frac{1}{\sqrt{n+1}})<2$