Question

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.

Answer 1

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.

Answer 2

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...

Answer 3

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.

Answer 4

Let us abbreviate the vector $(x,y,z,t)$ as $\mathbb{x}$. Combining Hölder's inequality and Young's inequality, $$ \|\mathbb{x}\|_3^{3/2} \leq \|\mathbb{x}\|_4 \|\mathbb{x}\|_2^{1/2} \leq \frac{3}{4} \|\mathbb{x}\|_4^{4/3} + \frac{1}{4}\|\mathbb{x}\|_2^2. $$ We do not have equality in the first inequality, because the entries of $\mathbb{x}$ are positive. Therefore $$ \|\mathbb{x}\|_3^{3/2} < \frac{3}{4} \|\mathbb{x}\|_4^{4/3} + \frac{1}{4}\|\mathbb{x}\|_2^2. $$ 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: $$ \|\mathbb{x}\|_3^{3/2} \leq \|\mathbb{x}\|_4^{4/3} \|\mathbb{x}\|_1^{1/6}, $$ where we do not have equality as before, and $\|\mathbb{x}\|_1=1$ by assumption.