Question

Answer 1

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(A=\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}=\frac{x^2}{x\sqrt{1-x^2}}+\frac{y^2}{y\sqrt{1-y^2}}\)

\(\geq \frac{(x+y)^2}{x\sqrt{1-x^2}+y\sqrt{1-y^2}}=\frac{1}{x\sqrt{1-x^2}+y\sqrt{1-y^2}}=\frac{1}{x\sqrt{y(2x+y)}+y\sqrt{x(2y+x)}}\)

Áp dụng BĐT Bunhiacopxky và AM-GM:

\([x\sqrt{y(2x+y)}+y\sqrt{x(2y+x)}]^2\leq (x^2y+y^2x)(2x+y+2y+x)=3xy(x+y)^2\)

\(\leq 3(\frac{x+y}{2})^2(x+y)^2=\frac{3}{4}\) 

$\Rightarrow x\sqrt{y(2x+y)}+y\sqrt{x(2y+x)}\leq \frac{\sqrt{3}}{2}$

$\Rightarrow A\geq \frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$ (đpcm)

Dấu "=" xảy ra khi $x=y=\frac{1}{2}$