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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
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4 Answers

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
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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
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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

                 (||x||_4^(4/3)-||x||_2^2)/(||x||_4^(4/3)-||x||_3^(3/2)) 

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

(||x||_1^(2/3)||x||_4^(4/3)-||x||_2^2)/(||x||_1^(2/3)||x||_4^(4/3)-||x||_1^(1/2)||x||_3^(3/2))

(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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