The Cuboid must be a Cube?

The relation can be written as

$\dfrac{a^3 + b^3 + c^3}{3} = abc.$

However, by the AM-GM inequality, we have

$\dfrac{a^3 + b^3 + c^3}{3} \geq \sqrt[3]{a^3b^3c^3} = abc.$

Note that $a, b, c$ are positive real numbers due to them being the side lengths of cubes, so AM-GM can be applied here.

Since we have equality, the side lengths must satisfy the equality conditions of AM-GM, meaning that $a^3 = b^3 = c^3,$ or $a = b = c.$ Thus, it is true that Babu’s cuboid is necessarily a cube because its side lengths are all the same.

We start by translating the given relation "The sum of the volumes of the cubes is three times the volume of the cuboid" into the following mathematical equation.

$a^3+b^3+c^3=3abc$

Now, $a^3+b^3+c^3=3abc$

$\implies a^3+b^3+c^3-3abc=0$

$\implies (a+b)^3-3ab(a+b)+c^3-3abc=0$

$\implies (a+b)^3+c^3-3ab(a+b)-3abc=0$

$\implies (a+b+c)\{(a+b)^2-(a+b)c+c^2\}-3ab(a+b+c)=0$

$\implies (a+b+c)(a^2+2ab+b^2-ca-bc+c^2-3ab)=0$

$\implies a^2+b^2+c^2-ab-bc-ca=0$ [As $a+b+c \neq 0$ ]

$\implies 2a^2+2b^2+2c^2-2ab-2ac-2ca=0$

$\implies (a-b)^2+(b-c)^2+(c-a)^2=0$

$\implies (a-b)^2=(b-c)^2=(c-a)^2=0$

$\implies a-b=b-c=c-a=0$

$\implies a=b=c$ .

So, $\boxed{\text{Yes}}$ , Babu's cuboid is necessarily a cube.

a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-bc-ca-ab) = 0, The first bracket cannot be zero as a, b and c are positive. The second bracket = {(b-c)^2+(c-a)^2+(a-b)^2}/2, which is zero only if a=b=c

This can be shown by a visual proof of the first octant, where $a,b,c$ are coordinates $x,y,z$ .

$f(x,y,z) = x^3 + y^3 + z^3$ is concave with respect to the origin; the farthest euclidean distance from the origin on a level surface $f(x,y,z)=k$ is achieved when $x=y=z$ .

$g(x,y,z) = 3xyz$ is convex with respect to the origin; the closest euclidean distance to the origin on a level surface $g(x,y,z)=k$ is achieved when $x=y=z$ .

Thus for the point $(x,y,z)$ , if $x=y=z$ , the level surfaces of $f$ and $g$ containing that point would have the same $k$ value, so $k=f(x,y,z)=g(x,y,z)$ .

But if $x=y=z$ were not true, the point $(x,y,z)$ would be on different level surfaces of $f$ and $g$ , meaning $f(x,y,z) \neq g(x,y,z)$ .

$(a + b + c)^3 = 3(a^2 + b^2 + c^2)(a + b + c) + 6abc - 2(a^3 + b^3 + c^3);$ with the given assumption, the last two terms cancel; divide out $a + b + c$ to obtain $(a + b + c)^2 = 3(a^2 + b^2 + c^2).$ Expand the left-hand side and reduce to $ab + bc + ac = a^2 + b^2 + c^2.$ This can be further reduced to $(c-a)(b-a) + (c - b)^2 = 0.$ We are free to rearrange so that $a \leq b \leq c$ , making the two terms on the left non-negative. But then both $(c-a)(b-a)$ and $(c-b)^2$ must be zero; thus $\begin{cases} c - b = 0 \\ \text{either}\ c-a = 0\ \text{or}\ b-a = 0,\end{cases}$ showing that $\boxed{a = b = c}$ .

Ordering a, b, c so c>=b>=a:

Let j=b-a>=0, k=c-a>=0

3abc =3a(a+j)(a+k) =3(a^3+ja^2+ka^2+ajk)

a^3+b^3+c^3 =a^3+(a+j)^3+(a+k)^3 =3a^3+3ja^2+3aj^2+j^3+3ka^2+3ak^2+k^3

a>0 so j=k=0 so a=b=c

If $3abc = a^3 + b^3 + c^3$ and if $a = x$ , then $x^3 - 3bcx + b^3 + c^3 = 0$ . Then graphically, all solutions for a positive real number $a = x$ would include all positive real x-intercepts of $y = x^3 - 3bcx + b^3 + c^3$ , where $b > 0$ and $c > 0$ .

The local maximum and minimum points of $y = x^3 - 3bcx + b^3 + c^3$ would be when $y’ = 3x^2 - 3bc = 0$ , or at $x = \pm \sqrt{bc}$ . Testing $x = -2\sqrt{bc}$ , $x = 0$ , and $x = 2\sqrt{bc}$ shows that this graph has a local maximum at $x = -\sqrt{bc}$ and a local minimum at $x = \sqrt{bc}$ . The local maximum height is therefore

$y = (-\sqrt{bc})^3 - 3bc(-\sqrt{bc}) + b^3 + c^3$

$y = -bc\sqrt{bc} + 3bc\sqrt{bc} + b^3 + c^3$

$y = 2bc\sqrt{bc} + b^3 + c^3$

$y = b^3 + 2bc\sqrt{bc} + c^3$

$y = (b\sqrt{b})^2 + 2(b\sqrt{b})(c\sqrt{c}) + (c\sqrt{c})^2$

$y = (b\sqrt{b} + c\sqrt{c})^2$

which is necessarily positive if b and c are real numbers.

Similarly, the local minimum height is

$y = (\sqrt{bc})^3 - 3bc(\sqrt{bc}) + b^3 + c^3$

$y = bc\sqrt{bc} - 3bc\sqrt{bc} + b^3 + c^3$

$y = -2bc\sqrt{bc} + b^3 + c^3$

$y = b^3 - 2bc\sqrt{bc} + c^3$

$y = (b\sqrt{b})^2 - 2(b\sqrt{b})(c\sqrt{c}) + (c\sqrt{c})^2$

$y = (b\sqrt{b} - c\sqrt{c})^2$

which is either $0$ if $b\sqrt{b} - c\sqrt{c} = 0$ or otherwise positive. If the local minimum height is positive, then for $x > 0$ the graph’s height decreases to a positive local minimum height and then increases, which means there are no positive x-intercepts, or no positive real solutions for $a$ . If the local minimum height is $0$ , then for $x > 0$ the graph’s height decreases to a local minimum height of $0$ and then increases, which means there is one positive x-intercept at the local minimum of $(\sqrt{bc}, b\sqrt{b} - c\sqrt{c})$ , where $b\sqrt{b} - c\sqrt{c} = 0$ . Solving this gives $b = c$ , and if $b = c$ then $x = a = \sqrt{bc} = b = c$ . Since this is the only possible positive real x-intercept, it is necessarily true that $a = b = c$ .