Evaluation of $f(a,b) = \min\left(\max\left(a^2+b\;,b^2+a\right)\right)$ 
Evaluation of $f(a,b) = \min\left(\max\left(a^2+b\;,b^2+a\right)\right)\;,$ Where $a,b\in \mathbb{R}$

$\bf{My\; Try::}$ First we have to calculate $\max(a^2+b,b^2+a) = \left\{\begin{matrix}
 a^2+b& \;,a^2+b>b^2+a \\\\ 
 b^2+a& \;,a^2+b\leq b^2+a \\ 
\end{matrix}\right.$
Now For $\bf{I^{st}}$ case, Here $f(a,b) = a^2+b\;\;,$If $ a^2+b>b^2+a\Rightarrow (a-b)\cdot (a+b-1)>0$
Similarly For $\bf{II^{st}}$ case, $f(a,b) = b^2+a\;\;,$If $ a^2+b\leq b^2+a\Rightarrow (a-b)\cdot (a+b-1)\leq 0$
Now I did not understand how can i solve it, Help me
Thanks 

I will also assume that you want to compute the quantity $Q:=\displaystyle\min_{(a,b) \in \mathbb{R}^2}\max(a^2+b,b^2+a)$
Notice that $\max(a^2+b,b^2+a)$ $\ge \dfrac{(a^2+b)+(b^2+a)}{2}$ $= \dfrac{1}{2}\left(a^2+a+b^2+b\right)$ $= \dfrac{1}{2}\left(a^2+a+\dfrac{1}{4}+b^2+b+\dfrac{1}{4}\right)-\dfrac{1}{4}$ $= \dfrac{1}{2}\left(a+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\left(b+\dfrac{1}{2}\right)^2-\dfrac{1}{4} \ge -\dfrac{1}{4}$. 
Can you figure out for what values of $a$ and $b$ do both of the inequalities become equalities?
