Weird inequalities!

Applying AM-GM inequality, we have:

$a^2+4\ge 2\sqrt{4a^2}=4a$

$b^3+b^3+\dfrac{512}{27}\ge 3\sqrt[3]{b^3.b^3.\dfrac{512}{27}}=8b^2$

$c^4+c^4+c^4+81\ge4\sqrt[4]{c^4.c^4.c^4.81}=12c^3$

So, $6\left(a^2+4\right)+3\left(b^3+b^3+\dfrac{512}{27}\right)+2\left(c^4+c^4+c^4+81\right)\ge 24(a+b^2+c^3)=\dfrac{2600}{3}$

Or $a^2+b^3+c^4\ge\dfrac{2807}{27}$

The equality holds when $a=2; b=\dfrac{8}{3}; c=3$ .

So $P+Q=2807+27=\boxed{2834}$ .

Lagrange multipliers give $b=\frac{4a}{3},c=\frac{3a}{2}$ . From the constraint we deduce that $a=2,b=\frac{8}{3},c=3$ , so that the minimum value is $\frac{2807}{27}$ , with $P+Q=\boxed{2834}$ . We can quickly check the boundary, where one or two of the variables are zero, using Lagrange multipliers once again.