# Prove $\left(\frac{a+b+c}{3}\right)^p\leq \frac{a^p+b^p+c^p}{3}$.

Prove: Let $$p$$ be an integer greater than $$1$$. Suppose $$a,b,c$$ be positive real numbers. Then $$\left(\frac{a+b+c}{3}\right)^p\leq \frac{a^p+b^p+c^p}{3}$$.

By AM-GM, I get $$\frac{a+b+c}{3}\geq (abc)^{1/3}$$ and $$\frac{a^p+b^p+c^p}{3}\geq (a^pb^pc^p)^{1/3}$$. Then $$\left(\frac{a+b+c}{3}\right)^p\geq (abc)^{p/3}$$.

I get stuck in how to relate $$\left(\frac{a+b+c}{3}\right)^p$$ and $$\frac{a^p+b^p+c^p}{3}$$. What should I do? I also face the same problem on proving $$\left(\frac{a+b+c+d}{4}\right)^p\leq \frac{a^p+b^p+c^p+d^p}{4}$$.

My tutor suggests me to let $$u=\frac{a+b}{2}$$ and $$v=\frac{c+d}{2}$$. I know $$\left(\frac{u+v}{2}\right)^p\leq \frac{u^p+v^p}{2}$$. So I substitute $$u$$ and $$v$$ into the inequality, and get $$\left(\frac{a+b+c+d}{4}\right)^p\leq \frac{{(a+b)}^p+{(c+d)}^p}{2^{p+1}}$$.

I haven't learnt Jensen's inequality. I suppose I should make use of AMGM to solve the problem(?

• Mar 21, 2022 at 6:17
• @MartinR Thank you, but I haven't learnt Jensen's inequality:( I cannot use it to solve the problem. Mar 21, 2022 at 6:21
• More general case: When $a_i$ are positive real numbers, then $(\frac{\sum_i a_i^x}{n})^{1/x}$ is non-decreasing function of $x$, which is constant if all $a_i$ are equal and increasing when not all $a_i$ are equal. Mar 21, 2022 at 9:16

To make the problem less tedious, let $$x = \dfrac{a}{a+b+c}, y = \dfrac{b}{a+b+c}, z = \dfrac{c}{a+b+c}$$, then you prove: $$x^p+y^p+z^p \ge \dfrac{1}{3^{p-1}}$$, with $$x+y+z=1$$. Let $$u = 3x, v = 3y, w = 3z$$, then you again prove: $$u^p+v^p+w^p \ge 3$$, with $$u+v+w = 3$$. Observe that by AM-GM inequality: $$u^p + 1(p-1) \ge p\sqrt[p]{u^p}=pu$$. Repeat this twice for $$v,w$$ we have: $$v^p + 1(p-1) \ge p\sqrt[p]{v^p}=pv$$, and $$w^p+1(p-1) \ge p\sqrt[p]{w^p}=pw$$. Adding these inequalities we obtain: $$u^p+v^p+w^p+3p-3 \ge p(u+v+w) = 3p\implies u^p+v^p+w^p \ge 3$$. Equality occurs when $$u = v = w = 1 \implies x = y = z \implies a = b = c$$. Done !