Cho các số thực dương x,y,z thỏa mãn: [tex]\sqrt{x}+\sqrt{y}+\sqrt{z}=1[/tex]
CMR:[tex]\sqrt{\frac{xy}{x+y+2z}}+\sqrt{\frac{yz}{y+z+2x}}+\sqrt{\frac{zx}{z+x+2y}}\leq \frac{1}{2}[/tex]
( áp dụng: [tex](a+b)^{2}\leq 2(a^{2}+b^{2})[/tex]
Theo BĐT Cauchy ta có:
[tex]\frac{x}{z+x}+\frac{y}{z+y}\geq 2\sqrt{\frac{x}{z+x}.\frac{y}{z+y}}=\frac{4\sqrt{xy}}{2\sqrt{(z+x)(z+y)}}\geq \frac{4\sqrt{xy}}{z+x+z+y}=\frac{4\sqrt{xy}}{x+y+2z}\\\Rightarrow \frac{\sqrt{xy}}{4}.\left ( \frac{x}{z+x}+\frac{y}{z+y} \right )\geq \frac{xy}{x+y+2z}\\\Rightarrow \sqrt{\frac{xy}{x+y+2z}}\leq \sqrt{\frac{\sqrt{xy}}{4}.\left ( \frac{x}{z+x}+\frac{y}{z+y} \right )}[/tex]
Tương tự....
Suy ra [tex]\sqrt{\frac{xy}{x+y+2z}}+\sqrt{\frac{yz}{y+z+2x}}+\sqrt{\frac{zx}{z+x+2y}}\leq \sqrt{\frac{\sqrt{xy}}{4}.\left ( \frac{x}{z+x}+\frac{y}{z+y} \right )}+\sqrt{\frac{\sqrt{yz}}{4}\left ( \frac{y}{x+y}+\frac{z}{x+z} \right )}+\sqrt{\frac{\sqrt{zx}}{4}.\left ( \frac{z}{y+z}+\frac{x}{y+x} \right )}(1)[/tex]
Ta có BĐT phụ: [tex](a+b+c)^2\geq 3(ab+bc+ca)[/tex] với [TEX]a,b,c>0[/TEX]
Áp dụng BĐT phụ trên ta được:
[tex]3(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})\leq (\sqrt{x}+\sqrt{y}+\sqrt{z})^2=1[/tex]
Theo BĐT Bunyakovsky ta có:
[tex]\left [ \sqrt{\frac{\sqrt{xy}}{4}.\left ( \frac{x}{z+x}+\frac{y}{z+y} \right )}+\sqrt{\frac{\sqrt{yz}}{4}\left ( \frac{y}{x+y}+\frac{z}{x+z} \right )}+\sqrt{\frac{\sqrt{zx}}{4}.\left ( \frac{z}{y+z}+\frac{x}{y+x} \right )} \right ]^{2}\\\leq \left ( \frac{\sqrt{xy}}{4}+\frac{\sqrt{yz}}{4}+\frac{\sqrt{zx}}{4} \right )\left [ \left ( \frac{x}{z+x}+\frac{y}{z+y} \right )+\left ( \frac{y}{x+y}+\frac{z}{x+z} \right )+\left ( \frac{z}{y+z}+\frac{x}{y+x} \right )\right ]\\= \left ( \frac{\sqrt{xy}}{4}+\frac{\sqrt{yz}}{4}+\frac{\sqrt{zx}}{4} \right ).3\\ =\frac{1}{4}.3(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})\\\leq \frac{1}{4}\\\Rightarrow \sqrt{\frac{\sqrt{xy}}{4}.\left ( \frac{x}{z+x}+\frac{y}{z+y} \right )}+\sqrt{\frac{\sqrt{yz}}{4}\left ( \frac{y}{x+y}+\frac{z}{x+z} \right )}+\sqrt{\frac{\sqrt{zx}}{4}.\left ( \frac{z}{y+z}+\frac{x}{y+x} \right )}\leq \frac{1}{2}(2)[/tex]
Từ (1) và (2) suy ra đpcm
Dấu = xảy ra khi x=y=z=1
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
