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

Question

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(?

Answer 1

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 !