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I encountered a problem today:

Prove that:

$$\frac{a^3+b^3+c^3}{a^2+b^2+c^2} \ge \frac{a+b+c}{3}$$

for all $a,b,c>0$

I used the RMS-AM inequality to replace the LHS with

$$\frac{\sqrt{a^2+b^2+c^2}}{\sqrt{3}}$$

and replaced the RHS using AM-GM inequality $$\frac{3abc}{a^2+b^2+c^2}$$

I can prove the new inequality, but does that mean I have proved the original inequality? I couldn't find another way to prove the original inequality except for expanding the terms and using scalar products. Thanks in advance!

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  • $\begingroup$ Do you have conditions upon a, b and c? Because for b=c=0 and a=-1 it doesn't seem to hold, does it? $\endgroup$ – ysearka May 3 '18 at 12:59
  • $\begingroup$ oh sorry, a,b,c>0 ill edit it inside the question $\endgroup$ – SuperMage1 May 3 '18 at 13:01
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We can use the Chebyshov's inequality.

Since $(a^2,b^2,c^2)$ and $(a,b,c)$ have the same ordering, we obtain: $$a^3+b^3+c^3=a^2\cdot a+b^2\cdot b+c^2\cdot c\geq\frac{1}{3}(a^2+b^2+c^2)(a+b+c).$$ Also, you can use PM and C-S.

Indeed, by PM $$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}\geq\sqrt{\frac{a^2+b^2+c^2}{3}},$$ which gives your $$\frac{a^3+b^3+c^3}{a^2+b^2+c^2}\geq\sqrt{\frac{a^2+b^2+c^2}{3}}.$$ Thus, it's enough to prove that $$\sqrt{3(a^2+b^2+c^2)}\geq a+b+c,$$ which is true by C-S: $$\sqrt{3(a^2+b^2+c^2)}=\sqrt{(1+1+1)(a^2+b^2+c^2)}\geq\sqrt{(a+b+c)^2}=a+b+c.$$

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  • $\begingroup$ I have solved the inequality like that, but is the method on the post an acceptable one by replacing the inequalities? $\endgroup$ – SuperMage1 May 3 '18 at 13:52
  • $\begingroup$ @SuperMage1 I added something. See now. By the way I think your $\frac{3abc}{a^2+b^2+c^2}$ gives a wrong inequality. Try $c\rightarrow0^+$, $a=b=1$. $\endgroup$ – Michael Rozenberg May 3 '18 at 14:25
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If $a,b,c>0$ the sequence $\{M_n = a^n+b^n+c^n\}_{n\geq 0}$ is log-convex by the Cauchy-Schwarz inequality, since the function $x\mapsto a^x+b^x+c^x$ is continuous and midpoint-log-convex.
In particular $M_3 M_0\geq M_1 M_2$.

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It is equivalent to $$(a^2-b^2)(a-b)+(a^2-c^2)(a-c)+(b-c)(b^2-c^2)\geq 0$$ and this is true.

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Note that \begin{align*} a^3+b^3+c^3 &= \frac{a^4}{a}+\frac{b^4}{b}+\frac{c^4}{c}\\ &\ge \frac{(a^2+b^2+c^2)^2}{a+b+c}. \end{align*} Then \begin{align*} \frac{a^3+b^3+c^3}{a^2+b^2+c^2} &\ge \frac{a^2+b^2+c^2}{a+b+c}\\ &\ge \frac{a+b+c}{3}, \end{align*} given that \begin{align*} (a+b+c)^2 \le 3(a^2+b^2+c^2). \end{align*}

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