Câu hỏi của Trần Anh Thơ

Question

Cho a,b,c là các số dương tùy ý. CMR $\frac{\sqrt{ab}}{c+2\sqrt{ab}}+\frac{\sqrt{bc}}{a+2\sqrt{bc}}+\frac{\sqrt{ca}}{b+2\sqrt{ca}}\le1$

Nguyễn Việt Lâm · Accepted Answer

Để dễ nhìn, đặt $\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)$

$VT=\frac{xy}{z^2+2xy}+\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}$

$2VT=\frac{2xy}{z^2+2xy}+\frac{2yz}{x^2+2yz}+\frac{2zx}{y^2+2xz}=1-\frac{z^2}{z^2+2xy}+1-\frac{x^2}{x^2+2yz}+1-\frac{y^2}{y^2+2xz}$

$2VT=3-\left(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\right)$

$2VT\le3-\frac{\left(x+y+z\right)^2}{x^2+2yz+y^2+2xz+z^2+2xy}=3-\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=2$

$\Rightarrow VT\le1$

Dấu "=" xảy ra khi $x=y=z$ hay $a=b=c$Câu trả lời: Để dễ nhìn, đặt $\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)$

$VT=\frac{xy}{z^2+2xy}+\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}$

$2VT=\frac{2xy}{z^2+2xy}+\frac{2yz}{x^2+2yz}+\frac{2zx}{y^2+2xz}=1-\frac{z^2}{z^2+2xy}+1-\frac{x^2}{x^2+2yz}+1-\frac{y^2}{y^2+2xz}$

$2VT=3-\left(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\right)$

$2VT\le3-\frac{\left(x+y+z\right)^2}{x^2+2yz+y^2+2xz+z^2+2xy}=3-\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=2$

$\Rightarrow VT\le1$

Dấu "=" xảy ra khi $x=y=z$ hay $a=b=c$

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