Proving algebraically $a^2+b^2\ge a^{\alpha}b^{2-\alpha}$ for $0\le\alpha\le2$ and $a,b\ge0$

Turning my comment into an answer: I don't know whether you'll think this is cheating, but assuming without loss of generality that $a\geq b$, then $$a^2+b^2 \geq a^2 = a^\alpha a^{2-\alpha} \geq a^\alpha b^{2-\alpha}. $$