Spring 2018, problem 58

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.

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.

• let M(a,b)=max(α,β) where α=3a^2+2b and β=3b^2+2*a.
• First solve for α>=β which yields a set S(say) {a>=b intersection a+b>=2/3 } union {a<=b intersection a+b<=2/3}.
• Our domain is now decreased!!
• I mean that every value M(α,β) will be    covered in this particular set S (i should say a part of domain) of a and b, because other set L(say)(rest part of domain)consisting of β>=α is just mirror image of set S if we plot a and b on x and y axis respectively.And off course The value M(α,β) will also be same for mirror images.[3a^2+2b for (a,b)€S and 3b^2+2 a for (a,b)€L because in above cases just the value of a and b are interchanged.] *** Note that now we just have to minimize α for set (domain) S.
• Let m=3a^2+2b...now This is a parabola Whose equation is y=(m/2)-(3/2)x^2.
• We have to minimize m (right??) so putting x=0 in above equation we get y=m/2.Which means that We have to decrease y coordinate(of a point where parabola cuts y axis) as much as possible.Plot the set S on coordinate axis with a and b on the x and y axis respectively and plot the parabola.But one should see that it should not be out of domain(I mean that choose minimal of y coordinate in such a way that the corresponding values of a and b should be in the domain S.)So after plotting This parabola and set S, we will come to know that the minimum value of m will occur when the line y=x will be tangent to the parabola y=(m/2)-(3/2)x^2 ...which will give m= -(1/3) which will give (3x+1)^2=0 ...giving x=y=-(1/3). It is solution that has never supposed anything !!(like other)Thanks for reading.You can ask by replying this answer if i was unable to satisfy you.

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 $a=b=\frac13$.