We start with a general cuboid with dimensions L, B and H (for Length, Breadth and Height) and introduce the letters P, S and V for Perimeter, Surface area and Volume where P=4(L+B+H), S=2(LB+BH+LH), and V=LBH. We will use the AGM (Arithmetic-Geometric Mean) inequality a number of times in the following.

First, using the AGM Inequality on L+B+H, we get L + B + H≥3(LBH)^{1/3 }with equality iff L=B=H. This is equivalent to (P/12)^{3}≥V.

Next, using the AGM Inequality on LB+BH+LH, we get LB+BH+LH≥3(L^{2}B^{2}H^{2})^{1/3 }with equality iff LB=BH=LH iff L=B=H.

This is equivalent to (S/6)^{3}≥V^{2}.

If we combine these inequalities by multiplying both left sides and both right sides we get PS/72≥V and it is easy to show that this is also with equality iff L=B=H.

Now, (S/2)^{2}=L^{2}B^{2}+B^{2}H^{2}+L^{2}H^{2}+2(L+B+H)V=L^{2}B^{2}+B^{2}H^{2}+L^{2}H^{2}+PV/2.

Also, (LB-BH)^{2}+(BH-LH)^{2}+(LH-LB)^{2}≥0 with equality iff L=B=H. Multiplying this out gives 2(L^{2}B^{2}+B^{2}H^{2}+L^{2}H^{2})≥PV/2.

Hence, (S/2)^{2}≥PV/4+PV/2=3PV/4.

From the above we get S^{2}≥3PV and yet again with equality iff L=B=H.

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