Có: \(\left(a^2+b^2\right)\left(a^5+b^5\right)=a^5+b^5+a^2b^3+a^3b^2\)
\(\Leftrightarrow\)\(a^2+b^2=\frac{a^2b^2\left(a+b\right)}{a^5+b^5}+1=\frac{a^2b^2\left(a+b\right)}{a^3+b^3}+1=\frac{a^2b^2}{a^2-ab+b^2}+1\le ab+1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=1\)