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If $V$ is the volume of a cuboid, $S$ is its total surface area, and $a, b,\; \text{and}\; c$ are the areas of its sides, then we have to prove that

${1\over{V^2}}=\frac{2}{S}(\frac1{a}+\frac1b+\frac1c)$

I think that $S$ must be twice the sum of $a, b,\; \text{and}\; c$, but now what?

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    It would be better if you used MathJax, and added a *please*. – 2017-01-31
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    But how to do ? – 2017-01-31
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    Read a Meta Post about it. – 2017-01-31

1 Answers 1

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Let the length of edges be $x,y,z$.

The area of the sides are: $a=xy,b=yz,c=xz$

The volume is $V=xyz$

The TSA is $S=2(xy+yz+xz)$

Insert these into your equation.

${1\over{(xyz)^2}}=({2\over{2(xy+yz+xz)}})({{xy+yz+xz}\over{(xyz)^2}})$

Which simplifies to

${1\over{(xyz)^2}}={1\over{(xyz)^2}}$

Thus proved.