2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $n\geq1$.

$\endgroup$
0
$\begingroup$

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 $n>0$.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ are you sure the last step is OK? ($n=1, 1.5\sqrt{3}>2.5\times 3?$) $\endgroup$ – chenbai May 31 '14 at 5:29
0
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ $a+b+c=\sqrt{3}\neq \dfrac{\sqrt{3}}{9} = abc$ $\endgroup$ – Math.StackExchange May 30 '14 at 1:29
  • $\begingroup$ Yes sorry. Can't see another example that works so I'll leave it trivial $\endgroup$ – John Fernley May 30 '14 at 11:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.