Proving cube volume larger than box volume with a constant surface area. Apologies for what may likely be a simply question. 
Given a surface area of $150$, my class was asked to find the maximum volume of a "box" with this surface area. This process was relatively simple:
$150 = 4s + 2s^2$, solved for $l$, then put into a volume equation, yielding:
$2V = 75s - 3s^2$
$s = 5$
This took my precalc class an unnecessary amount of time since were all still in very basic math. I figured that the max volume of any "box" is always a cube (and that this could be applied to any other three dimensional shape), so you could skip the prior steps by simply dividing your surface area by the amount of sides on the shape, then taking the square root of that number, yielding the length of one side of your shape. In the case of a cube:
$$\sqrt\frac{SA}{6}$$
Visual Representation: https://www.desmos.com/calculator/j872cerb6z
In order to use this I have to prove it works, and this is where I am struggling. My thought is to prove that a cubes volume is always greater than that of a non-cube "box" when they have equal surface areas:
Cube SA = Box SA 
$$6s^2 = 2lw + 2lh + 2wh$$
$$s = \sqrt\frac{lw+lh+wh}{3}$$
Volume Cube - Volume Box $=$ 
$$= \sqrt\frac{lw+lh+wh}{3} - 2lw + 2lh + 2wh$$
Showing that this works out to be positive would show that a cube always has a larger volume than a non-cube box with the same surface area, but I attempting to take it any further seems beyond my capabilities. Any help with proving (or disproving) this would be appreciated.
 A: I think we're better off showing it a different way.  First, we show that a square has a smaller perimeter than any other rectangle of the same area.  Consider a rectangle of sides $s+t$ and $s-t$, where $s > t > 0$.  Its area is $s^2-t^2$.  A square of the same area would have a side of $\sqrt{s^2-t^2} < \sqrt{s^2} < s$, so its perimeter would be smaller than that of the rectangle.
Now, we move to the three-dimensional case.  Suppose we had a rectangular solid of length $l$, width $w$, and depth $d$, with $l > w$.  Consider the cross-section of area $lw$.  We claim that a rectangular solid of greater volume can be generated by using a square cross-section of side $s$.  By the above lemma, we know that we can create a cross-section of equal area $s^2 = lw$, but with perimeter $4s < 2(l+w)$.  Thus a solid of dimensions $s$ by $s$ by $d$ would have the same volume as the original solid, but less surface area.  By the same token, however, we could make the solid deeper than $d$, to restore the original surface area, but increase the volume.
By repeating the same argument with width and depth (instead of length and width), we establish that a rectangular solid of whatever dimensions always has a smaller volume than a cube of identical surface area.
The above argument is far from rigorous, but it can be made so, and maybe it's convincing enough for your purposes?
