How to prove this for positive real numbers? $$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$

I tried AM-GM, CS inequality but all failed.

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    $\begingroup$ I hope it is given that $a,b,c \gt 0$. $\endgroup$ – Indrayudh Roy Feb 17 '14 at 14:42
  • $\begingroup$ Why somebody downvote? $\endgroup$ – zhangwfjh Feb 17 '14 at 14:44
  • $\begingroup$ Does $a$, $b$, $c$ are arbitrary real numbers? Or they are positive numbers? $\endgroup$ – Lion Feb 17 '14 at 14:50
  • $\begingroup$ @Lion Non zero reals. $\endgroup$ – zhangwfjh Feb 17 '14 at 14:52
  • $\begingroup$ Do you have try the method of extreme value of multivariable function? $\endgroup$ – Lion Feb 17 '14 at 14:54

Using Cauchy-Schwarz Inequality twice:

$a^4 + b^4 +c^4 \geq a^2b^2 +b^2c^2 +c^2a^2 \geq ab^2c +ba^2c +ac^2b = abc(a+b+c)$

  • $\begingroup$ Substitute $a,b,c$ in $a^2 + b^2 +c^2 \geq ab +bc +ca$ with $ab,bc,ca$ respectively. $\endgroup$ – r9m Feb 17 '14 at 14:49
  • $\begingroup$ If $abc<0$ then you get the reverse inequality, right? $\endgroup$ – user37238 Feb 17 '14 at 14:54
  • $\begingroup$ $a^2 + b^2 +c^2-ab-bc-ca = 1/2((a-b)^2 +(b-c)^2 +(c-a)^2) \geq 0$ so it holds for all $a,b,c$ $\endgroup$ – r9m Feb 17 '14 at 15:00
  • $\begingroup$ Please try $a=-2$, $b=1$ and $c=1$. $\endgroup$ – user37238 Feb 17 '14 at 15:02
  • $\begingroup$ @user129017, when you divide both sides by $abc$, then the inequality gets reversed when $abc \lt 0$. $\endgroup$ – Indrayudh Roy Feb 17 '14 at 15:03

I have come up with an answer with myself. Using CS inequality $$(a^4+b^4+c^4)(1+1+1)\geq(a^2+b^2+c^2)^2$$ $$(a^2+b^2+c^2)(1+1+1)\geq(a+b+c)^2$$ Hence we have $$a^4+b^4+c^4\geq\frac{(a+b+c)^4}{27}=(a+b+c)\left(\frac{a+b+c}{3}\right)^3\geq abc(a+b+c)$$

  • $\begingroup$ Your answer is invalid. $$\left(\frac{a+b+c}{3}\right)\ge abc$$ does not hold for all $a,b,c\in\mathbb R$. Mean inequalities are only used for positive numbers (except for $\frac{a+b}{2}\ge \sqrt{ab}$). If you still don't believe me, take, e.g., $a=-1$, $b=-1$ and $c=1$. $\endgroup$ – user26486 Feb 22 '14 at 0:51

Here other two answers used Cauchy-Scwartz Inequality. I am giving a simple $AM\ge GM$ inequality proof.

You asked, $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge a+b+c\\\implies a^4+b^4+c^4\ge a^2bc+b^2ca+c^2ab$$

Now, from, $AM\ge GM$, we have $$\frac {a^4+ a^4+b^4+c^4}4\ge \left(a^4\cdot a^4\cdot b^4\cdot c^4\right)^{1/4}=a^2bc\tag 1$$

Similarly, $$\frac {a^4+ b^4+b^4+c^4}4\ge \left(a^4\cdot b^4\cdot b^4\cdot c^4\right)^{1/4}=ab^2c\tag 2$$ and also, $$\frac {a^4+ b^4+c^4+c^4}4\ge \left(a^4\cdot b^4\cdot c^4\cdot c^4\right)^{1/4}=abc^2\tag 3$$

Now, summing up $(1),(2),(3)$, we have, $a^4+b^4+c^4\ge a^2bc+b^2ca+c^2ab$, that is $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge a+b+c$$


By Holder $$\sum_{cyc}\frac{a^3}{bc}\geq\frac{(a+b+c)^3}{3(ab+ac+bc)}=\frac{(a+b+c)\cdot(a+b+c)^2}{3(ab+ac+bc)}\geq a+b+c$$


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