\(\Leftrightarrow\frac{a^2+b^2}{2}-ab-\frac{\left(a-b\right)^2}{a^2+b^2+2}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2}{2}-\frac{\left(a-b\right)^2}{a^2+b^2+2}\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(\frac{1}{2}-\frac{1}{a^2+b^2+2}\right)\ge0\)
\(\frac{\Leftrightarrow\left(a-b\right)^2\left(a^2+b^2\right)}{2\left(a^2+b^2+2\right)}\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(a=b\)
Có một cách khác nè :3 Nhưng đương nhiên vẫn dài hơn cách anh Lâm :v
\(ab+\frac{\left(a-b\right)^2}{a^2+b^2+2}=\frac{a^3b+ab^3+2ab+a^2-2ab+b^2}{a^2+b^2+2}\)
\(=\frac{a^2\left(ab+1\right)+b^2\left(ab+1\right)}{a^2+b^2+2}\le\frac{\left(ab+1\right)\left(a^2+b^2\right)}{2ab+2}\left(vi-a^2+b^2\ge2ab\right)\)
\(=\frac{\left(ab+1\right)\left(a^2+b^2\right)}{2\left(ab+1\right)}=\frac{a^2+b^2}{2}=VT\)
Done :3