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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

4 Answers 4

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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At x=y=z=t and at x=1 (two extreme cases) the quotient is not defined. otherwise (up to permutation) WLOG we may assume that x leq y leq z leq t.

one idea is to assume that x increases and t decreases (by the same small amount), leaving y,z intact. this makes the problem a one-parameter maximization problem. by examining the first order optimization condition Grad=0 i imagine that some conclusion can be made, whether x should be increased or decreased, etc etc.

i can only guess that the extreme cases cited above provide global maxima and minima.

in doing the calculations i advise you to use a symbolic software...

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typically, "answers" are full answers, suggestions for how one might arrive at an answer are more appropriate posted as a "comment" to the question –  Carlo Beenakker Sep 19 '12 at 16:18

Another approach is to use inequalities for p-norms which can be found in text books of normed spaces, advanced linear algebra or functional analysis. basically the question is to bound


given that ||x||_1=1.

one problem in the above expression is its lack of homogeneity. this can be corrected by replacing the above expression e.g. by


(using the fact that the change only involves multiplication by 1) where both numerator and denominators are now 2-homogeneous and so the condition ||x||_1=1 may be dropped. i can just assume that this last expression should come out from the usual bounds which relate these norms. namely, an upper bound for the numerator and a lower bound for the denominator.

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