# Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$

Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.

I know the proof by subtracting LHS by RHS and then doing some arrangement.

But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an easy proof?

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Look up Muirhead's inequality if that's what you are thinking about. – Soarer Jul 3 '11 at 14:46
Why do a, b, and c have to be positive and unequal? – Peter Olson Jul 3 '11 at 16:57
@Peter: otherwise the inequality may not be strict. – Qiaochu Yuan Jul 3 '11 at 19:46

The result is also a consequence of the Chebyshev (sum) Inequality. This is a quite useful result, not the least for contest problems! It can be proved by using the Rearrangement Inequality, which is also useful to know. For completeness, we state the full result, though only half of it is needed here.

Theorem: (Chebyshev Inequality) Suppose that $a_1 \le a_2 \le \cdots \le a_n$ and $b_1 \le b_2 \le \cdots \le b_n$.

Let $m=a_1b_n + a_2b_{n-1} +\cdots + a_nb_1$ and $M=a_1b_1+a_2b_2+\cdots + a_nb_n$. Then $$nm \le (a_1+a_2+\cdots+a_n)(b_1+b_2+\cdots +b_n) \le nM$$ with equality only happening when all the $a_i$ are equal or all the $b_i$ are equal.

To apply the result to our problem, we can without loss of generality assume that $a \le b \le c$. Let $a_1=a$, $a_2=b$, $a_3=c$, $b_1=a^2$, $b_2=b^2$, and $b_3=c^2$.

Then the conditions of Chebyshev's Inequality are met, and $M=a^3+b^3+c^3$. We can then read off our inequality from the Chebyshev Inequality.

Note that the Chebyshev Inequality yields an immediate generalization of the $3$ variable inequality of the question to $n$ variables.

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As is often the case, we can use Cauchy-Schwarz.

Since the $L^3$ norm is larger then the $L^4$ norm, we have that $$\left(a^{3}+b^{3}+c^{3}\right)^{\frac{1}{3}}\geq\left(a^{4}+b^{4}+c^{4}\right)^{\frac{1}{4}}$$ with equality if and only if $a=b=c.$ Using Cauchy Schwarz, we have that $$\sqrt{3\left(a^{4}+b^{4}+c^{4}\right)}\geq\left(a^{2}+b^{2}+c^{2}\right)$$ and $$\sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}\geq\left(a+b+c\right).$$ Multiplying these two inequalities, we get $$3\left(a^{4}+b^{4}+c^{4}\right)^{\frac{3}{4}}\geq\left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}\right),$$ which implies the result.

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You can express the inequality as

$$\frac{a^3+b^3+c^3}{3} > \frac{a+b+c}{3} \frac{a^2+b^2+c^2}{3}$$

or

$$[m_3(a,b,c)]^3 > [m_1(a,b,c)] \; [m_2(a,b,c)]^2$$

where $m_p$ is the generalized power (Hölder) $p$-mean. Applying the (important) property that $m_p > m_q$ whenever $p>q$ (and the values are not all equal), you're done.

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There is a fairly general way to prove inequalities like this using rearrangement, but just for fun, let me show you how you can prove a large class of inequalities like this by repeated use of AM-GM. First subtract $a^3, b^3, c^3$ from the RHS, then symmetrically sum the inequalities

$$a^3 + a^3 + b^3 \ge 3 a^2 b.$$

Essentially I am using weighted AM-GM with rational coefficients. Here are some exercises that can be solved using this idea.

Exercise 1. Prove that if $n > 1$ is a positive integer and $a > 0$, then

$$\frac{1 + a + ... + a^n}{a + a^2 + ... + a^{n-1}} \ge \frac{n+1}{n-1}.$$

Exercise 2. Let $a, b, c \ge 0$. Prove that

$$a^5 + b^5 + c^5 \ge 5abc(b^2 - ac).$$

(Weighted AM-GM isn't necessarily the best way to solve either of these problems; it's just the argument I came up with first.)

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