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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.

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Can you link to the question at the other site? And have you put a link there to this site? – Gerry Myerson Sep 19 '12 at 12:56
I linked the question to Math Stack Exchange. – ivan Sep 19 '12 at 14:26

2 Answers 2

up vote 8 down vote accepted

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.

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Very nice multiplications by $s_1$! – ivan Sep 20 '12 at 6:19
@ivan: You only need the assumption $s_1=1$ to verify that the denominator is positive. Everything else follows without this assumption, see my response. – GH from MO Sep 20 '12 at 7:54
good point! I was multiplying by $s_1$ to homogenize, and somehow I didn't realize it was unnecessary... – zeb Sep 20 '12 at 8:30
@zeb: I did the same, and only realized at the end that it was not necessary! – GH from MO Sep 20 '12 at 13:33

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.

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