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I ran into a following problem in The Cauchy-Schwarz Master Class:

Let $x, y, z \geq 0$ and $xyz = 1$.

Prove $x^2 + y^2 + z^2 \leq x^3 + y^3 + z^3$.

The problem is contained in the chapter about symmetric polynomials and Muirhead's inequality.

The proof based on Muirhead's inequality is pretty quick:

We multiply the left hand side with $\sqrt[3]{xyz} = 1$ and prove $$x^{\frac{7}{3}}y^{\frac{1}{3}}z^{\frac{1}{3}} + x^{\frac{1}{3}}y^{\frac{7}{3}}z^{\frac{1}{3}} + x^{\frac{1}{3}}y^{\frac{1}{3}}z^{\frac{7}{3}} \leq x^3 + y^3 + z^3$$

with Muirhead ( $(3, 0, 0)$ majorizes $(\frac{7}{3}, \frac{1}{3}, \frac{1}{3})$).

I'm curious if there's a way to prove this without machinery of Muirhead's inequality and majorization. Also, this approach readily generalizes to proving

$$x^n + y^n + z^n \leq x^{n+1} + y^{n+1} + z^{n+1}$$

for non-negative $x, y, z$ such that $xyz = 1$. Is there a way to prove this generalization without Muirhead?

Community
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ante.ceperic
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  • Since Muirhead can be proven using inequalities like arithmetic mean $>$ geometric mean $>$ harmonic mean, you can prove this by copying Muirhead's proof with variable replaced by those in your inequality. – Ragnar Aug 19 '14 at 00:27
  • Yet another way would be to show that $f(x) = x^{n+1}-x^n-\log x$ is non-negative, which follows from checking the sign of $f'(x) = nx^n(x-1)+\frac1x(x^{n+1}-1)$. – Macavity Aug 19 '14 at 04:39

3 Answers3

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In general, using the power mean inequality,

$$\sqrt[n+1]{\frac{x^{n+1} + y^{n+1} + z^{n+1}}{3}} \ge \sqrt[n]{\frac{x^n + y^n + z^n}{3}}$$

Raising both sides to the same power,

$$\frac{x^{n+1} + y^{n+1} + z^{n+1}}{3} \ge \frac{x^n + y^n + z^n}{3} \cdot \sqrt[n]{\frac{x^n + y^n + z^n}{3}}$$

But, again by the power mean inequality,

$$\sqrt[n]{\frac{x^n + y^n + z^n}{3}} \ge \sqrt[3]{xyz} = \sqrt[3]{1} = 1$$

Hence,

$$x^{n+1} + y^{n+1} + z^{n+1} \ge x^n + y^n + z^n$$

With equality at $x=y=z=1$.

Yiyuan Lee
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From Chebyshev's sum inequality we have \begin{align*} \frac{x^{n+1}+y^{n+1}+z^{n+1}}{3}\geq \frac{x^n+y^n+z^n}{3}\cdot\frac{x+y+z}{3}. \end{align*} By AM-GM we have $\frac{x+y+z}{3}\geq\sqrt[3]{xyz}=1$ and that proves the desired inequality.

S.B.
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Another approach, not using the general power inequalities.

For convenience, we say $a=x^{\frac 13}$ and similar for $b$ and $c$.

Just apply the inequality arithmetic mean $>$ geometric mean in the following way (and cyclic permutations): \begin{align} \frac 79 a^9+\frac 19 b^9+\frac 19 c^9&=\frac{\underbrace{a^9+a^9+\cdots+a^9}_{\text{7 times}}+b^9+c^9}{9}\\ &\geq \sqrt[9]{\left(a^9\right)^7 b^9c^9}=a^7bc \end{align} Summing the cyclic permutations gives the required inequality.
In the general version, use the following: (where $a=x^{\frac 13}$ again) \begin{align} \frac{\underbrace{a^{3n+3}+a^{3n+3}+\cdots+a^{3n+3}}_{\text{3n+1 times}}+b^{3n+3}+c^{3n+3}}{3n+3} &\geq \sqrt[b^{3n+3}]{\left(a^{3n+3}\right)^{3n+1} b^{3n+3}c^{3n+3}}=a^{3n+1}bc \end{align}

Ragnar
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