\(\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+2-2\sqrt{ }x+1}=\frac{x+5}{2}\)\(\frac{x+5}{2}\)
Xét \(x^2+\sqrt{1+x^2}\)ta có:
\(x^2\ge0\)
nên \(1+x^2\ge1\)
\(\Rightarrow\sqrt{1+x^2}\ge\sqrt{1}=1\)
\(\Rightarrow x^2+\sqrt{1+x^2}\ge1\)
Tương tự ta có:
\(y^2+\sqrt{1+y^2}\ge1\)
Do đó: \(\left(x^2+\sqrt{1+x^2}\right)\left(y^2+\sqrt{1+y^2}\right)\ge1\)
Dấu bằng xảy ra khi \(x=0;y=0\)
Khi đó \(x+y=0\left(ĐPCM\right)\)