Cho $a,b,c> 0$.CMR :
$\sum \sqrt{\frac{a^2}{a^2+7ab+b^2}}\geq 1$
Cho $a,b,c> 0$.CMR :
$\sum \sqrt{\frac{a^2}{a^2+7ab+b^2}}\geq 1$
Ta có :$\sum \sqrt{\frac{a^2}{a^2+7ab+b^2}}=\sum \frac{1}{\sqrt{1+\frac{7b}{a}+(\frac{b}{a})^2}}$
Đặt $\frac{b}{a}=m,\frac{c}{b}=n,\frac{a}{c}=p= > mnp=1$ .Đặt $m=\frac{yz}{x^2},n=\frac{xz}{y^2},p=\frac{xy}{z^2}$
$= > \sum \frac{1}{\sqrt{m^2+7m+1}}=\sum \frac{1}{\sqrt{(\frac{yz}{x^2})^2+\frac{7yz}{x^2}+1}}=\sum \frac{x^2}{\sqrt{x^4+7yzx^2+y^2z^2}}=\sum \frac{x^4}{x.\sqrt{x^6+7yzx^4+x^2y^2z^2}}\geq \frac{(\sum x^2)^2}{\sum x\sqrt{x^6+7yzx^4+y^2z^2x^2}}\geq \frac{(\sum x^2)^2}{\sqrt{(\sum x^2)(\sum x^6+3x^2y^2z^2+7xyz(\sum x^3))}}\geq 1< = > (\sum x^2)^4\geq (\sum x^2)(\sum x^6+3x^2y^2z^2+7xyz(\sum x^3))< = > 3\sum x^4(y^2+z^2)+3x^2y^2z^2\geq 7xyz(\sum x^3)$(Luôn đúng theo AM-GM )
đã có ở đây
Cho $a,b,c> 0$.CMR :
$\sum \sqrt{\frac{a^2}{a^2+7ab+b^2}}\geq 1$
http://diendantoanho...caa2geqslant-1/
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