I [asked this question](https://math.stackexchange.com/questions/194666/power-means-inequality) at Stack Exchange but received no answer. The origins of the question are unclear, as I came across it rummaging through old notebooks from highschool, in one of which it was stated without any reference or proof. Let $x, y, z$ and $t$ be positive numbers such that $x+y+z+t=1$. Then the following inequality holds:
$$
\frac{\sqrt[3]{x^4 + y^4 + z^4 + t^4} - (x^2 + y^2 + z^2 + t^2)}{\sqrt[3]{x^4 + y^4 + z^4 + t^4} - \sqrt{x^3 + y^3 + z^3 + t^3}}<4
$$


I tried various approaches, e.g. using some form of power means monotonicity, symmetric reduction, looking up Bullen's "Handbook of Means and Their Inequalities", even desperate approaches like the y-positivity of Cuttler, Greene & Skandera. It didn't work. 


I doubt this is a research grade question even though numerical experiments show that it can be extended to any number of variables, not only 4. Moreover, I believe that an elementary proof exists otherwise I would not have been able to prove it in highschool.



An inequality involving sums of powers
Let $s\_k = x^k + y^k + z^k + t^k$. First we check that the denominator is nonnegative. By Holder, we know $s\_4^{2/3}s\_1^{1/3} \ge s\_3$, rearranging that and using $s\_1 = 1$ we see that the denominator is indeed at least $0$.


Now we multiply out and rearrange, to see that the given inequality is equivalent to:


$\frac{3\sqrt[3]{s\_4s\_1^2}+s\_2}{4} \ge \sqrt{s\_3s\_1}$.


By the weighted AM-GM inequality the left hand side is at least $\sqrt[4]{s\_4s\_2s\_1^2}$, so the inequality boils down to showing that $s\_4s\_2 \ge s\_3^2$, which follows from Cauchy-Schwartz.


Note that no step of this proof depends on the number of variables.


