Find the minimum of this expression This is a problem in my exam and I can't find the solution using elementary inequality knowledge. Can anyone here help me solve this. Thanks


$a,b,c $ are positive real numbers which satisfy $(a+c)(b+c) = 4c^{2}$. Find the minimum of this expression: 
    $$P = \frac{32a^{3}}{(b+3c)^{3}} + \frac{32b^3}{(a+3c)^{3}} - \frac{\sqrt{a^{2} + b^{2}}}{c}$$


Thanks so much.
 A: We begin with the observation  that the change $a=xc, b=yc$ reduces the problem under consideration to the two-dimensional optimization problem 
$$\min \frac {x^3} {(y+3)^3} + \frac {y^3} {(x+3)^3}-\sqrt{x^2+y^2}$$
under the constraints $$(x+1)(y+1)=4, x \ge 0, y \ge 0 . $$
Solving it with Mathematica by
Minimize[ 32 x^3/(y + 3)^3 + 32 y^3/(x + 3)^3 - Sqrt[x^2 + y^2], 
 (x + 1)*(y + 1) == 4 && x >= 0 && y >= 0, {x, y}]

we obtain:
{1 - \sqrt{2}, {x -> 1, y -> 1}}
i.e. $\{a=b,b=c\}.$ One may play with Lagrange multipliers to this end.

Addition. For each nonnegative $x$ and for each nonnegative $y$ the inequality
  $$ \frac {x^3} {(y+3)^3} + \frac {y^3} {(x+3)^3}-\sqrt{x^2+y^2} \ge 2\sqrt{\frac {x^3y^3} {(x+3)^3(y+3)^3}}-\sqrt{x^2+y^2}$$ holds.
  The equality takes place iff $\frac {x^3} {(y+3)^3} =\frac {y^3} {(x+3)^3}.$ This implies $x=y$ because the function $f(x):=x^3(x+3)^3$ increases for nonnegative $x$. Taking into account $(x + 1)(y + 1) =4$, we obtain the optimal solution $\{x=1, y=1\}$. 
