Non-numerical proof of an inequality 
Let $ 0 < s < 1 $ be the unique solution to the equation $$ \frac{1}{2^s} + \frac{1}{6^s} + \frac{1}{12^s}=1, $$ and show that $$ \frac{1}{6^s}+\frac{1}{12^s} \geq \left(\frac{2}{7}\right)^s. $$

It is easy do this numerically; you can compute $ s \simeq 0.7584 $ with a calculator, and show that the desired inequality holds with this value plugged in.
I'm looking for a more general proof, maybe geometric, using some calculus, or using concavity of the function $ x \mapsto x^s $ for $ 0 < s < 1 $.

Denote $a = \frac{1}{2^s}$ and $b = \frac{1}{3^s}$.
We have $a + a b + a^2 b = 1$ which results in $b = \frac{1 - a}{a + a^2}$.
Also, since $a^2 < b$, we have $\frac{1 - a}{a + a^2} > a^2$
or $\frac{(a^2 + 1)(1 - a - a^2)}{a + a^2} > 0$ which results in $1 - a - a^2 > 0$.
Since $x\mapsto x^s$ is concave on $x>0$, we have
$(\frac{6}{7})^s + (\frac{8}{7})^s \le 2(\frac{\frac{6}{7} + \frac{8}{7}}{2})^s = 2$
which results in $(\frac{2}{7})^s \le \frac{2}{4^s + 3^s}$.
It suffices to prove that $\frac{1}{6^s} + \frac{1}{12^s} > \frac{2}{4^s + 3^s}$ or
$ab + a^2 b > \frac{2}{\frac{1}{a^2} + \frac{1}{b}}$
or $\frac{(1-a)^2(a+1)(1 - a - a^2)}{a^4 + a^3 - a + 1} > 0$ which is true.
We are done.
