Chủ đề này có 3 trả lời

[quynhhph1]

Cho $\frac{1}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{z}+\sqrt{x}}=3$

Tìm max $A=\frac{1}{2\sqrt{x}+3\sqrt{y}+3\sqrt{z}}+\frac{1}{2\sqrt{y}+3\sqrt{x}+3\sqrt{z}}+\frac{1}{3\sqrt{x}+3\sqrt{y}+2\sqrt{z}}$

[kfcchicken98]

có $\frac{1}{2\sqrt{x}+3\sqrt{y}+3\sqrt{z}}=\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{x}+\sqrt{z}+\sqrt{y}+\sqrt{z}+\sqrt{y}+\sqrt{z}}\leq \frac{1}{16}(\frac{1}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{x}+\sqrt{z}}+\frac{1}{\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{y}+\sqrt{z}})=\frac{3}{16}+\frac{1}{16(\sqrt{y}+\sqrt{z})}$

tương tự, $A\leq \frac{9}{16}+\frac{3}{16}=\frac{3}{4}$

[Phuong Thu Quoc]

Dưa vào Cauchy-Schwarz thôi $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\geq \frac{16}{a+b+c+d}$

Ta có $\frac{1}{2\sqrt{x}+3\sqrt{y}+3\sqrt{z}}=\frac{1}{\left ( \sqrt{x}+\sqrt{y} \right )+\left ( \sqrt{x} +\sqrt{z}\right )+\left ( \sqrt{y}+\sqrt{z} \right )+\left ( \sqrt{y}+\sqrt{z} \right )}\leq \frac{1}{16}.\left ( \frac{1}{\sqrt{x}+\sqrt{y}} +\frac{1}{\sqrt{x}+\sqrt{z}}+\frac{1}{\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{y}+\sqrt{z}}\right )$

Xây dựng các BĐT tương tự ta được $A\leq \frac{1}{4}\left ( \frac{1}{\sqrt{x}+\sqrt{y}} +\frac{1}{\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{y}+\sqrt{z}}\right )=\frac{3}{4}$

[Simpson Joe Donald]

$A \leq \dfrac{1}{16}(\dfrac{4}{\sqrt{x}+\sqrt{y}}\dfrac{4}{\sqrt{x}+\sqrt{z}}+\dfrac{4}{\sqrt{z}+\sqrt{y}})=\dfrac{3}{4}$

 

Dấu $"="$ xảy ra khi và chỉ khia $x=y=z=0,25$