Any neat proof that $0$ is the unique solution of the equation $4^x+9^x+25^x=6^x+10^x+15^x$?

Question

It is obvious that both $f(x)= 4^x+9^x+25^x$ and $g(x)=6^x+10^x+15^x$ are strictly monotonic increasing functions. It is also easy to check that $0$ is a solution of the equation.

Also I chart the functions, and it looks that for any $x$, $f(x)>g(x),$ which can be somehow proof by studying the derivative of the $h(x)=f(x)-g(x)$ and showing that $(0,0)$ is an absolute minimum point for $h(x).$

However $h(x)$ is a function with a messy derivative, and is not looking easy (for me) to find the zeroes of this derivative.

Does anyone know an elegant proof (maybe an elementary one, without derivatives) for this problem?

Answer 1

HINT: $$a^2+b^2+c^2\geq ab+bc+ca$$

Answer 2

Since the question is already settled for the continuous version, I thought I would just add (for fun!) a short proof for the case in which $x$ is a whole number: for $x \geq 1$, the two sides are not equal when reduced modulo ten. So, the only possible whole number solution is at $x=0$, which works. "QED"