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

My Analysis professor showed this inequality and elegantly proved it using polar coordinates, saying that it can't be done algebraically. Instead, here's how I think I have handled it: firstly we see it's true for $ab=0$; dividing by the RHS we get $$\left(\frac{a}{b}\right)^{2-\alpha}+\left(\frac{a}{b}\right)^{-\alpha}\ge1,$$ or equivalently, setting $t=a/b$, $$t^2+1\ge t^\alpha$$ which holds because the LHS is $\ge t^2\ge t^\alpha$ for $t\ge1$ and $\ge1\ge t^\alpha$ for $0\le t<1$.

Am I missing something? Are there fancy algebraic (perhaps some linear algebra inequalities) ways to prove the inequality?

• It can't be done by... is always a dangerous thing to say in maths :) – Jack D'Aurizio Oct 3 '16 at 21:20
• +1 for not being discouraged by prof. However, maybe a hardcore algebraists won't even like exponentiation with irrational exponent? – Hagen von Eitzen Oct 3 '16 at 21:21
• @JackD'Aurizio: Indeed, ahahah :D – Vincenzo Oliva Oct 3 '16 at 21:25
• Another way: Write $\alpha = 1 + \beta$ with $\beta \in [-1,1]$. Divide by $ab$ (after disposing of the trivial case $ab = 0$) to get $$\frac{a}{b} + \frac{b}{a} \geqslant \biggl(\frac{a}{b}\biggr)^{\beta}.$$ – Daniel Fischer Oct 3 '16 at 21:33
• Or, if you assume that $a\geq b$, then $$a^2+b^2\geq a^2 = a^\alpha a^{2-\alpha} \geq a^\alpha b^{2-\alpha}.$$ – Pierre-Guy Plamondon Oct 3 '16 at 21:36

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 $$a^2+b^2 \geq a^2 = a^\alpha a^{2-\alpha} \geq a^\alpha b^{2-\alpha}.$$
Ask him to prove, using polar coordinates, that $$a^2+b^2\ge ka^{\alpha}b^{2-\alpha}\quad \forall 0\le\alpha\le 2, \forall a,b\ge 0$$ where $$k=\frac{2}{\left(\alpha^{\alpha}(2-\alpha)^{(2-\alpha)}\right)^{1/2}}.$$ (Which is the best constant possible.)