From Square to Cube

• Let $a, b$ and $c$ denote the dimensions of the rectangular cuboid. Then $\sqrt{a^2+b^2+c^2}$ represents the diagonal length while the volume is expressed by $abc$ .

• Of course, $a^2,b^2,c^2>0$ . By the superb AM-GM inequality, $\dfrac{a^2+b^2+c^2}{3} \geq \sqrt[3]{a^2b^2c^2} \text{, the equality holds when } a^2=b^2=c^2 \text{ or equivalently, } a=b=c$ $\Leftrightarrow \dfrac{(\sqrt{a^2+b^2+c^2})^3}{3\sqrt{3}} \geq abc$

So, the max value of the volume occurs when $a=b=c$ , that is, when the cuboid is a cube.

Hence, $\boxed{\text{True}}$ is the answer.

Let $a,b,c$ be the squares of the lengths of the sides. $D = a + b + c$ is the square of the diagonal, and $V = abc$ is the squared volume.

The constraint is $D = \text{constant}$ ; we wish to maximize $V$ . Taking the derivative, $0 = dD = da + db + dc;$ $dV = bc\:da + ac\:db + ab\:dc = bc\:da + ac\:db - ab\:(da + db) = b(c - a)\:da + a(c - b)\:db;$ This must be zero for changes in $a$ and $b$ independently; thus $b(c-a) = a(c-b) = 0$ . If $a,b \not= 0$ , we have $c - a = 0, c - b = 0\ \ \ \therefore\ \ \ a = b = c.$

This proves that at the maximum value of $V$ , the three sides are equal.