# 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

• How can you take the minimum of one number? – user137731 May 17 '15 at 4:50

(Update)

I assume that you want to compute the quantity $$Q:=\min_{(a,b)\in{\Bbb R}^2}\max\{a^2+b,b^2+a\}\ .$$ The difference $$(b^2+a)-(a^2+b)=(b-a)(b+a-1)$$ vanishes when $u:=b-a=0$, or when $v:=a+b-1=0$. This suggest that we introduce $u$ and $v$ as new variables, which leads to $$2a=1-u+v,\quad 2b=1+u+v\qquad\bigl((u,v)\in{\Bbb R}^2\bigr)\ .\tag{1}$$ One then computes $$4(a^2+b)=3+4v+(u-v)^2,\quad 4(b^2+a)=3+4v+(u+v)^2\ .$$ It follows that $$\max\{a^2+b,b^2+a\}={1\over4}\bigl(3+4v+\max\{(u-v)^2,(u+v)^2\}={1\over4}\bigl(3+4v+\bigl(|u|+|v|\bigr)^2\bigr)\ .$$ Now we have to minimize the right hand side over $(u,v)\in{\Bbb R}^2$. For given $v$ this is minimal when $u=0$, so that we are left with the task to minimize $$f(v):={1\over4}(3+4v+v^2)={1\over4}\bigl((v+2)^2-1\bigr)\ .$$ The minimum $-{1\over4}$ is taken when $v=-2$. Putting $u=0$, $v=-2$ in $(1)$ leads to $a=b=-{1\over2}$.

• How about $a=b=- \frac{1}{2}$? – ThePortakal May 17 '15 at 15:58
• @ThePortakal: Thank you for spotting my blunder. – Christian Blatter May 17 '15 at 17:22

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.

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?