Với mọi số thực x;y ta có \(\left(x-y\right)^2\ge0\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow2x^2+2y^2\ge x^2+2xy+y^2\)
\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2\)
\(\Rightarrow8\left(\left(a^2\right)^2+\left(b^2\right)^2\right)\ge4\left(a^2+b^2\right)^2\ge4\left(\frac{1}{2}\left(a+b\right)^2\right)^2=\left(a+b\right)^4\)
Dấu "=" có xảy ra khi \(a=b\)