Rectangular prisms!

This is actually a direct application of the Power Mean Inequality.

Let the height, width and length of the prism be $x,y,z$ respectively. The sum of all edge lengths is given as $2x+2y+2y+2z+2x+2z=4x+4y+4z$ , since there are four of each edge in the prism.

Since $4x+4y+4z =36$ , we get that $x+y+z = 9$ , and what we are asked for is the smallest possible length of a diagonal, which is $\sqrt{x^2+y^2+z^2}$ . By the power mean inequality, we get $\sqrt{\frac{x^2+y^2+z^2}{3}}\geq \frac{x+y+z}{3} = \frac{9}{3} = 3$ . Multiplying both sides by $\sqrt{3}$ gives us $\sqrt{x^2+y^2+z^2} \geq 3\sqrt{3}$ , so we get $a=b=3$ . Our answer is then $a+b = 3+3 = \boxed{6}$ .

My approach was not really the intended one, it uses the rearrangement inequality instead.

Let the width, length, and height of the prism be $x, y,$ and $z$ . Since $4x+4y+4z = 36$ , $x+y+z=9$ .

Square both sides to get $x^2+y^2+x^2+2(xy+yz+zx)=81$ .

Moving things, we get $xy+yz+zx=\frac{81 - (x^2+y^2+z^2)}{2}$ .

From the Rearrangement Inequality, $x^2+y^2+z^2 \geq xy + yz + zx$ .

So, $x^2+y^2+z^2 \geq \frac{81 - (x^2+y^2+z^2)}{2}$ .

$3(x^2+y^2+z^2) \geq 81$

$x^2+y^2+z^2 \geq 27$

$\sqrt{x^2+y^2+z^2} \geq \boxed{3\sqrt{3}}$

This is obtainable when $x=y=z=3$ .