Prove that $\left(\dfrac{1}{n}\sum_{i=1}^n x_i^5\right)^{1/5}\geq \dfrac{1}{n}\sum_{i=1}^n x_i$

Question

Show that if $x_1,x_2,...,x_n$ are positive numbers then $\left(\dfrac{1}{n}\sum_{i=1}^n x_i^5\right)^{1/5}\geq \dfrac{1}{n}\sum_{i=1}^n x_i$.

I'm a little stuck on this question. I think I might need to use convex functions. I know the left sum is equal to $\dfrac{1}{n^5}(\sum_{i=1}^n x_i^5)^{1/5}$. There are rules like $(a+b+c+d+\dots)^2 = a^3 + b^3 +c^3+d^3+\dots$ for consecutive integers $a,b,c,d,...$. This is very easy to show using induction, though I don't think induction will be very useful here.

edit: apparently Jensen's inequality could be useful.

Answer 1

Use Hölder's inequality: for nonnegative $x_i, y_i$ following holds $$ \sum_{i=1}^n x_i y_i \le \left( \sum_{i=1}^n x_i^p \right)^{1/p} \cdot \left( \sum_{i=1}^n y_i^{p^{'}} \right)^{1/p^{'}}, $$ where $$ \frac{1}{p} + \frac{1}{p^{'}} = 1, p>1. $$

One can take $y_i = 1$, $p = 5$ and get $$ \sum_{i=1}^n x_i \le \left( \sum_{i=1}^n x_i^5 \right)^{1/5} \left( \sum_{i=1}^n 1^{5/4} \right)^{4/5} = n \cdot \left( \frac{1}{n}\sum_{i=1}^n x_i^5 \right)^{1/5}. $$ Dividing both sides by $n$ gives the result $$ \frac{1}{n}\sum_{i=1}^n x_i \le \left( \frac{1}{n}\sum_{i=1}^n x_i^5 \right)^{1/5}. $$

Answer 2

Let $f(x)=x^5$, then f''(x)=20x^3>0, for x>0, so from Jensen's inequality ir follows that $$\frac{f(a)+f(b)+f(c)}{3} \ge f(\frac{a+b+c}{3}).$$ $$\frac{a^5+b^5+c^5}{3} \ge (\frac{a+b+c}{3})^{5}$$ and hence the result.

Answer 3

One may also use Tchebecheff's inequality repeatedly. If $a \ge b \ge c $ and $p \ge q \ge r$, then $$(ap+bq+cr) \ge \frac{(a+b+c)(p+q+r)}{3}~~~~(*)$$ We have $$(a^2+b^2+c^2) \ge \frac{(a+b+c)(a+b+c)}{3} \ge \frac{(a+b+c)^2}{3} ~~~(1)$$ Next $$(a^3+b^3+c^3) \ge \frac{(a+b+c)(a^2+b^2+c^2)}{3} \ge \frac{(a+b+c)^3}{9} ~~~(2.$$ Finally from (*), (1) and (2) we get $$(a^2~a^3+b^2~b^3+c^2~c^3)\ge \frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{3} \ge \frac{(a+b+c)^5}{81}.$$ Or $$\frac{a^5+b*5+c^5}{3} \ge \frac{(a+b+c)^5}{3^5}$$