When I read a book, the author said that it was easy to prove this result, but I thought about it for a long time and didn't prove it, so I ask it here. Thank you.
Problem: Under the assumption $0<a\le 1\le b$ fix $a^2b+b^2a$, then the minimal value of $a^2+ab+b^2$ occurs when $1\in\{a\}\bigcup\{b\}$ .
By AM-GM $$a^2+ab+b^2=\sqrt[3]{\frac{1}{8}(2a^2+2ab+2b^2)^3}=\frac{1}{2}\sqrt[3]{((a+b)^2+a^2+b^2)^3}\geq$$ $$\geq\frac{1}{2}\sqrt[3]{\left(2\cdot\frac{(a+b)^2}{2}+2ab\right)^3}\geq\frac{1}{2}\sqrt[3]{\left(3\sqrt[3]{\left(\frac{(a+b)^2}{2}\right)^2\cdot2ab}\right)^3}=$$ $$=\frac{3}{2}\sqrt[3]{\frac{(a+b)^4ab}{2}}\geq\frac{3}{2}\sqrt[3]{\frac{(a+b)^2\cdot4ab\cdot ab}{2}}=\frac{3}{\sqrt[3]4}\left(a^2b+b^2a\right)^{\frac{2}{3}}.$$ The equality occurs for $a=b$, which says that we got a minimal value.