4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Ragnar Aug 19 '14 at 0:27
  • $\begingroup$ 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 $f'(x) = nx^n(x-1)+\frac1x(x^{n+1}-1)$. $\endgroup$ – Macavity Aug 19 '14 at 4:39
4
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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 $x=y=z=1$.

$\endgroup$
3
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.