## Spring 2018, problem 58

Let $a$ and $b$ be two real numbers and let $M(a,b)=\max\left \{ 3a^{2}+2b; 3b^{2}+2a\right \}$. Find the values of $a$ and $b$ for which $M(a,b)$ is minimal.

### Comments

Any chance of expanding this answer with some explanatory notes ?

I'm probably not the only one who's not familiar with the terminology.

For example what does "AM-GM's" mean?

Thanks, Phil.

The function $M(a,b)$ is minimized when both of its arguments equal each other. That is:

$3a^2 + 2b = 3b^2 + 2a \Rightarrow 3a^2 - 3b^2 + 2b - 2a = 0 \Rightarrow 3(a+b)(a-b) - 2(a-b) = 0 \Rightarrow (a-b)[3(a+b) - 2] = 0$

or when: $a = b$ (i), $a+b = \frac{2}{3}$ (ii). Substituting (i) into either argument results in the quadratic equation:

$f(a) = 3a^2 + 2a$

whose first & second derivavtives yield:

$f'(a) = 6a + 2 = 0 \Rightarrow a = -\frac{1}{3}$;

$f''(-\frac{1}{3}) = 6 > 0$ (hence, a global minimum)

which produces the critical pair $P_{1}(a,b) = (-\frac{1}{3}, -\frac{1}{3})$. Likewise, substituting (ii) produces:

$g(a) = 3a^2 + 2(\frac{2}{3} - a) = 3a^2 - 2a + \frac{4}{3}$;

$g'(a) = 6a - 2 = 0 \Rightarrow a = \frac{1}{3}$;

$g''(\frac{1}{3}) = 6 > 0$ (also a global minimum)

and the critical pair $P_{2}(a,b) = (\frac{1}{3}, \frac{1}{3})$. Checking both critical pairs against $M(a,b)$ now results in:

$M(-\frac{1}{3}, -\frac{1}{3}) = 3(-\frac{1}{3})^2 + 2(-\frac{1}{3}) = \boxed{-\frac{1}{3}}$.

$M(\frac{1}{3}, \frac{1}{3}) = 3(\frac{1}{3})^2 + 2(\frac{1}{3}) = \boxed{1}$.

Since $M_{P_{1}} \le M_{P_{2}}$, $\boxed{a = b = -\frac{1}{3}}$ are the critical values.

Now this is a solution I can understand !

However, the rather bold statement at the beginning surely needs some explaining.

i.e. "The function M(a,b) is minimized when both of its arguments equal each other"

Phil.

Apart from the obvious mistake (a wrong choice between the two extremal points) this solution is identical to mine; perhaps more detailed, but still without a justification for the equal values of the functions at the minimum point. As explained in my reply to Hubert, if a function is bigger than the other one near a minimum point for $M(a, b)$, such a function should have a local minimum in such a point; however none of our functions can have local minima.

The minimality of $M(a,b)$ forces $3a^2+2b=3b^2+2a$, say $3a^2-2a=3b^2-2b$.

E.g. solving in $a$ we get $a=\frac{1\pm(3b-1)}3$ say $a=b$ or $a=\frac23-b$.

The first case bring to minimize $3b^2+2b$ and choosing $b=-\frac13$ we get the value $-\frac13$.

The second case asks for the minimum of $3b^2-2b+\frac43$ and choosing $b=\frac13$ we get the value 1.

Thus $\min M(a,b)=1$, reached with $a=b=\frac13$.

Check that $M(-\frac 13; -\frac 13)=-\frac 13$, and how do you show that minimality yields equality of the terms ?

**modified on february 11**

@Hubert: thank you for signalling my mistakes.

In particular I forget to say that functions must agree unless the minimum is reached in a local minimum of one of them.

My answer can be adjusted, but anyway your solution is better than mine

Call this $M=\text{max}\{u; v\}$,

now with $(s,p)=(a+b,ab)$, AM-GM's $\Rightarrow~4p\le s^2\Rightarrow~~$

$2M(a,b)\ge u+v=3(s^2-2p)+2s\ge \frac 32s^2+2s$

$=\frac 32(s+\frac 23)^2-\frac 23\ge -\frac 23=2M(-\frac 13; -\frac 13),$ the answer.