# Prove the inequality $a^n+b^n+c^n>3(1+\frac{n}{2})$ using induction

Given positive numbers $a,b,c$,

satisfying $a+b+c=abc$,

how to prove the following inequality

$$a^n+b^n+c^n>3\left(1+\frac{n}{2}\right)$$

Maybe this can be proved by induction.

by AM_GM inequality: $a + b + c = abc \leq \dfrac{(a + b + c)^3}{27} \Rightarrow (a + b + c)^2 \geq 27 \Rightarrow a + b + c\geq 3\sqrt{3} \Rightarrow (a + b + c)^n \geq 3^{\frac{3n}{2}}$.

But $f(x) = x^n$ is convex on $(0,\infty)$ for $n > 2$. Thus:

$f(a) + f(b) + f(c) \geq 3\cdot f\left(\dfrac{a+b+c}{3}\right)$. So:

$a^n + b^n + c^n \geq 3\cdot \left(\dfrac{a+b+c}{3}\right)^n = 3^{1-n}\cdot (a + b + c)^n \geq 3^{1-n}\cdot 3^{\frac{3n}{2}} = 3^{1 + \frac{n}{2}} > 3(1 +\frac{n}{2})$, whereas the last inequality can be proven by induction.

By the AM-GM Inequality, $$\frac{abc}{3}=\frac{a+b+c}{3}\geq \sqrt[3]{abc}.$$ Solving for $abc$, we get $abc\geq 3\sqrt{3}$. Therefore, $$\frac{a^n+b^n+c^n}{3}\geq \left(\frac{a+b+c}{3}\right)^n=\left(\frac{abc}{3}\right)^n\geq\left(\frac{3\sqrt{3}}{3}\right)^n=3^{\frac{n}{2}},$$ where the first inequality comes from Jensen's Inequality. Hence, $$a^n+b^n+c^n \geq 3^{1+\frac{n}{2}}.$$ This is what I think you meant to ask. However, it's also true that $a^n+b^n+c^n >3\left(1+\frac{n}{2}\right)$. To see this, observe that $$\frac{d}{dn}\left[3^{1+\frac{n}{2}}-3\left(1+\frac{n}{2}\right)\right]= \frac{3}{2}\left(3^{n/2}\ln{3}-1\right) \geq \frac{3}{2}(\sqrt{3}\ln{3}-1)>0$$ for all $n\geq1$. Therefore, since $3^{1+\frac{1}{2}}-3\left(1+\frac{1}{2}\right)=3\sqrt{3}-\frac{9}{2}>0$, it follows that $3^{1+\frac{n}{2}}>3\left(1+\frac{n}{2}\right)$ for all $n\geq1$.

Minimum of $a+b+c$ can be calculated using Lagrange multipliers, minimize$$a+b+c+\lambda(a+b+c-abc)$$the solution is $a = b = c = \sqrt 3$, so $a+b+c\ge 3^{3/2}$, then using Holder inequality, $$[a,b,c]\cdot[1,1,1]=a+b+c\le(a^n+b^n+c^n)^{1/n}\cdot3^{1-1/n}$$ so that is $$a^n+b^n+c^n\ge3^{1+n/2}>3+\frac{3\ln3}{2}n$$ last inequality is first order taylor expansion, I assum $n>0$.

using Chebyshev's sum inequality,

$$\begin{array}{lcr} a^na+b^nb+c^nc&>&a^nb+b^nc+c^na\\ a^na+b^nb+c^nc&>&a^nc+b^na+c^nb\\ a^na+b^nb+c^nc&=&a^na+b^nb+c^nc \end{array}$$

sum inequalities above, $$a^na+b^nb+c^nc>\frac{1}{3}(a+b+c)(a^n+b^n+c^n)>\sqrt{3}(1+\frac{n}{2})>3(1+\frac{n+1}{2})$$

• are you sure the last step is OK? ($n=1, 1.5\sqrt{3}>2.5\times 3?$) – chenbai May 31 '14 at 5:29

Note that $n$ needs to be a positive integer, e.g. $n=0$ fails (trivially)

• $a+b+c=\sqrt{3}\neq \dfrac{\sqrt{3}}{9} = abc$ – Math.StackExchange May 30 '14 at 1:29
• Yes sorry. Can't see another example that works so I'll leave it trivial – John Fernley May 30 '14 at 11:22