\(\left(x+1\right)\left(y+1\right)=4xy\Leftrightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=4\)
Đặt \(\left(\frac{1}{x};\frac{1}{y}\right)=\left(a;b\right)\Rightarrow\left(a+1\right)\left(b+1\right)=4\Rightarrow ab+a+b=3\)
\(VT=\frac{a}{\sqrt{a^2+3}}+\frac{b}{\sqrt{b^2+3}}\)
Áp dụng BĐT Bunhiacopxki:
\(\left(a+\sqrt{3}.\sqrt{3}\right)^2\le\left(1+3\right)\left(a^2+3\right)\Rightarrow a^2+3\ge\frac{\left(a+3\right)^2}{4}\)
\(\Rightarrow VT\le\frac{2a}{a+3}+\frac{2b}{b+3}=\frac{4ab+6\left(a+b\right)}{ab+3\left(a+b\right)+9}=\frac{4\left(ab+a+b\right)+2\left(a+b\right)}{ab+a+b+9+2\left(a+b\right)}=\frac{12+2\left(a+b\right)}{12+2\left(a+b\right)}=1\)
Dấu "=" xảy ra khi \(a=b=1\) hay \(x=y=1\)