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Let $x_1, \ldots, x_4$ be real numbers. I wish to show that there exists a constant $C$ such that

$$x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 \leq C(x_1^2 +x_2^2 + x_3^2 + x_4^2)^{3/2}$$

Moreover, I want to find the smallest such constant.

This must involve some identity with symmetric polynomials, but I haven't been able to get any to work. I've tried the usual (Newton, Muirhead, etc.)

user182973
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  • Hint: So long as you are not after the optimal $C$, it is enough to take a single term on the left. Then you get the similar inequality for the other terms by permuting the variables, and the final $C$ will be four times the one for a single term. Moreover, to prove it with $x_1x_2x_3$ on the left, you can drop the $x_4$ term on the right. Take the cube root. Do you recognize the inequality then? – Harald Hanche-Olsen Jun 24 '14 at 07:37
  • Another approach is to use homogeneity to restrict yourself to the case where $x_1^2+x_2^2+x_3^3+x_4^2=1$ and use compactness. But since you use the algebra-precalculus tag, I suspect you don't have know about compactness yet. – Harald Hanche-Olsen Jun 24 '14 at 07:41
  • Yes. Actually, you have a similar inequality between geometric mean and mean square which is more appropriate, but that follows from the AM-GM inequality applied to the squared terms. – Harald Hanche-Olsen Jun 24 '14 at 07:46
  • Thus, I get $x_1^2 x_2^2 x_2^3 ≤(x_2^1+x_2^2+x_2^3)^3$, and we take square roots. – user182973 Jun 24 '14 at 07:54
  • I think you mixed up some sub- and superscript. And you forgot to divide by 3 somewhere (the mean of three numbers, after all). But yes, you have got it. – Harald Hanche-Olsen Jun 24 '14 at 08:25

1 Answers1

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If $x_1=x_2=x_3=x_4=1$ we get $C\geq\frac{1}{2}$.

We'll prove that $C=\frac{1}{2}$ is valid.

We need to prove that: $$(x_1^2+x_2^2+x_3^2+x_4^2)^{\frac{3}{2}}\geq2( x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)$$ and since the left side does not depend on replacing $x_i\rightarrow-x_i$,

it's enough to prove the last inequality for non-negative variables.

In another hand, $$x_1^2+x_2^2+x_3^2+x_4^2\geq\frac{2}{3}(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)$$ it's just $$(x_1-x_2)^2+(x_1-x_3)^2+(x_1-x_4)^2+(x_2-x_3)^2+(x_2-x_4)^2+(x_3-x_4)^2\geq0.$$ thus, it remains to prove that $$\left(\frac{2}{3}(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)\right)^{\frac{3}{2}}\geq$$ $$\geq2( x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)$$ or $$\sqrt{\frac{x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4}{6}}\geq\sqrt[3]{\frac{x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4}{4}}$$ and see here: How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$

Done!

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Michael Rozenberg
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