An inequality involving sums of powers I asked this question 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.
 A: 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.
A: Let us abbreviate the vector $(x,y,z,t)$ as $\mathbf{x}$. Combining Hölder's inequality and Young's inequality,
$$ |\mathbf{x}|_3^{3/2} \leq |\mathbf{x}|_4 |\mathbf{x}|_2^{1/2} 
\leq \frac{3}{4} |\mathbf{x}|_4^{4/3} + \frac{1}{4}|\mathbf{x}|_2^2. $$
We do not have equality in the first inequality, because the entries of $\mathbf{x}$ are positive. Therefore
$$ |\mathbf{x}|_3^{3/2} < \frac{3}{4} |\mathbf{x}|_4^{4/3} + \frac{1}{4}|\mathbf{x}|_2^2. $$
Rearranging, we obtain the desired inequality. In this last step we use that the denominator is positive, which is another application of Hölder's inequality: 
$$ |\mathbf{x}|_3^{3/2} \leq |\mathbf{x}|_4^{4/3} |\mathbf{x}|_1^{1/6}, $$
where we do not have equality as before, and $|\mathbf{x}|_1=1$ by assumption.
