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 

you can go on further, for example, 1st case:
sub case 1: $a \ge b, a+b \ge 1$ we need to find min$a^2+b$
$a^2+b=a^2-a+a+b \ge a^2-a+1=(a-\dfrac{1}{2})^2+\dfrac{3}{4} \ge \dfrac{3}{4} $ when $a=b=\dfrac{1}{2}$ get min.
rest cases you can do simliar thing.
