<s>[INST] Solving an infimum-inequality How does one evaluate $$\inf \{c: c^2 \geq \frac{a^2x^2+b^2y^2}{x^2+y^2} \} $$ for every $a,b,x,y \in \mathbb R$ ? 
Hints and tips appreciated
Edit: I tried to write the inequality in polar coordinates, which gives me:
$$ \frac{a^2r^2\cos^2(\phi) + b^2r^2\sin^2(\phi)}{r^2\cos^2(\phi) + r^2\sin^2(\phi) } \leq c^2$$ thus
$a^2\cos^2(\phi) + b^2\sin^2(\phi) \leq c^2 $
Since $\sin$ and $\cos$ are bounded by $1$, I receive $a^2+b^2 \leq c^2 $
Is this correct $( \color{blue}{really}$ the infimum and not just some upper bound?) 
Edit: Without loss of generality, let $a^2 \leq b^2 $. Then I can find a lower upper bound by estimating $a^2\cos^2(\phi) + b^2\sin^2(\phi) \leq c^2$ to $b^2 \leq c^2$, thus $c = \max\{ \lvert a\rvert\, \lvert b \rvert \}$
 [/INST] It's $c = \sqrt{\max(a^2, b^2)}$. Proof: 


*

*If $c = \sqrt{\max(a^2, b^2)}$, then $a^2 \leqslant c^2$ and $b^2 \leqslant c^2$, so 


$$a^2 \cos^2( \phi ) + b^2 \sin^2( \phi ) \leqslant c^2 \cos^2( \phi ) + c^2 \sin^2( \phi ) = c^2.$$


*If $\gamma^2 \geqslant a^2 \cos^2( \phi ) + b^2 \sin^2( \phi )$ for all $x, y \in \mathbb{R}$, then letting $\phi = 0$ and $\phi = \frac{\pi}{2}$ we get $\gamma^2 \geqslant a^2$ and $\gamma^2 \geqslant b^2$, so $\gamma \geqslant c$.


Note: I assume you want $c$ positive, because very large negative $c$ will always satisfy the inequality. 
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