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vutienhoang

vutienhoang

Đăng ký: 25-02-2015
Offline Đăng nhập: 15-09-2016 - 20:13
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Trong chủ đề: CMR: $\frac{1}{a}+\frac{2}...

14-08-2015 - 10:18

$(a^2+2b^2)(\frac{1}{a}+\frac{2}{b})^2\geq (1+2)^3=3c^2(\frac{3}{c})^2$ ( holer)
có a^2+b^2$\leq$ 3c^2 =>$\frac{1}{a}+\frac{2}{b}\geq \frac{3}{c}$
 


Trong chủ đề: $P=\frac{x^2y}{z^3}+\frac{y^2z...

13-08-2015 - 21:48

P=$\sum \frac{x^2y^3}{y^2z^3}+\frac{4}{\sum \frac{x}{y}}$
đặt $\frac{x}{y}=a ; \frac{y}{z}=b ; \frac{z}{x}=c => abc=1$
P trở thành $a^2b^3+b^2c^3+c^2a^3+\frac{4}{a+b+c}$
Có $(ab+bc+ac)^2\geq 3abc(a+b+c)$ (bđt cơ bản)
<=>$(ab+bc+ac)^2\geq 3(a+b+c) =>\frac{4}{a+b+c}\geq \frac{12}{(ab+bc+ac)^2}$
Có $a^2b^3+b^2c^3+c^2a^3=a^2b^2b+b^2c^2c+c^2a^2a=\frac{a^2b^2}{ac}+\frac{b^2c^2}{ab}+\frac{c^2a^2}{bc}\geq \frac{(ab+bc+ac)^2}{ab+bc+ac}=ab+bc+ac$
=> P$\geq ab+bc+ac+\frac{12}{(ab+bc+ac)^2}$


Trong chủ đề: P= $\frac{2+a}{\sqrt{2-a}}+...

12-08-2015 - 11:07

P=$\frac{2+a}{\sqrt{1+b}}+\frac{2+b}{\sqrt{1+a}}$=$\frac{1}{3}(\frac{1+a}{\sqrt{1+b}}+\frac{1+b}{\sqrt{1+a}})+\frac{2}{3}\frac{1+a}{\sqrt{1+b}}+\frac{2(1+b)}{3\sqrt{1+a}}+\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}\geq \frac{1}{3}(\frac{(2+a+b)^2}{\sqrt{(1+a)(1+b)}(\sqrt{1+a}+\sqrt{1+b})})+4\sqrt{\frac{2}{3}}\geq \frac{3}{(\frac{a+b+2}{2})(\sqrt{1+a}+\sqrt{1+b})}+4\sqrt{\frac{2}{3}}$
có$\sqrt{\frac{3}{2}(1+a)}+\sqrt{\frac{3}{2}(1+b)}\leq \frac{\frac{5}{2}+a+\frac{5}{2}+b}{2}=3 =>\sqrt{1+a}+\sqrt{1+b}\leq 3\sqrt{\frac{2}{3}}$


Trong chủ đề: $a,b,c\in R.a^2+b^2+c^2=3.Max:a+b+c-abc$

09-08-2015 - 11:04

 đặt A =a+b+c-abc
$A^{2}\leq (a^2+1)((bc-1)^2+(b+c)^2)=(a^2+1)(b^2+1)(c^2+1)\leq (\frac{a^2+b^2+c^2+3}{3})^3=8$
<=>$A\leq 2\sqrt{2}$
 


Trong chủ đề: chứng minh rằng $\sum \frac{a^{3}}...

06-08-2015 - 23:07

$\sum \frac{a^3}{b+c}=\sum \frac{a^4}{ab+bc}\geq \frac{(\sum a^2)^2}{2\sum ab}\geq \frac{\sum a^2}{2}$
=> $3\sum \frac{a^3}{b+c}\geq \sum \frac{a^3}{b+c}+\sum a^2=\sum \frac{a^2(a+b+c)}{b+c}$
=> ta cần cm $\sum \frac{a^2}{b+c}\geq \sum \frac{a}{b+c}. (\frac{a+b+c}{3})=\sum \frac{a^2}{3(b+c)}+\sum \frac{b+c}{3} <=> \frac{2a^2}{3(b+c)}\geq \frac{2(a+b+c)}{3}$  ( đúng theo cauchy-schwarz)