Never Judge a Problem by its Title

$P\left( x \right) =3{ \left( x+\cfrac { 2 }{ 3 } \right) }^{ 2 }+\cfrac { 5 }{ 3 } \\ \\ \Rightarrow P\left( x \right) \ge \cfrac { 5 }{ 3 } \\ \\ \& \quad Q\left( y \right) =3{ \left( y+1 \right) }^{ 2 }+12\\ \\ \Rightarrow Q\left( y \right) \ge 12\\ \\ \therefore \quad P\left( x \right) .Q\left( y \right) \ge \cfrac { 5 }{ 3 } .12\\ \\ \Rightarrow P\left( x \right) .Q\left( y \right) \ge 20,so\\ \\ P\left( x \right) .Q\left( y \right) =20\quad \quad only\quad if\quad x=\cfrac { -2 }{ 3 } ,y=-1\\ \\ \Rightarrow P\left( \cfrac { -2 }{ 3 } \right) +Q\left( -1 \right) =\cfrac { 5 }{ 3 } +12=\cfrac { 41 }{ 3 } \\ \\ \Rightarrow a+b=41+3=44$

A possibly more mechanical solution is letting $k$ be a certain number such that:

$3x^2 + 4x + 3 = k$

$3y^2 + 6y + 15 = \frac{20}{k}$

You want the first and the second equations to have real roots, so you use the discriminant of both ( $\ge 0$ ) and find that the only possible value of $k$ is $\frac{5}{3}$ . Thus,

$3x^2 + 4x + 3 = \frac{5}{3}$

$3y^2 + 6y + 15 = 12$

Examining both equations, we see that $P(x) = \frac{5}{3}$ and $Q(y) = 12$ . Their sum is $\frac{41}{3}$ so the answer is $41 + 3 = \boxed{44}$ .

[3 (x + 2/ 3)^2 + 5/ 3][3 (y + 1)^2 + 12] = 20

Only occurs with x = - 2/ 3 and y = -1 while both are at minimum for (5/ 3)(12) = 20, therefore 5/ 3 + 12 = 41/ 3.

41 + 3 = 44