Prove that for every three positive numbers $a, b, c$:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \ge \frac{a+b+c}{a+b+c+\sqrt[3]{abc}}$$

I tried using $\sum_{\mathrm{cyc}}$ but I haven't got far. I also tried: $$\sqrt[3]{abc}=\frac{1}{3}\sqrt[3]{3a \cdot 3b \cdot 3c}$$ and according to the HM-GM inequality we can replace the root by: $$\frac13 \frac{3}{\frac1{3a} +\frac1{3b}+\frac1{3c}}=\frac{1}{\frac13\left(\frac1a+\frac1b+\frac1c\right)}=\frac{3}{ \frac1a+\frac1b+\frac1c}$$ And that's it.

Can you give me a hint or a solution for the question?

  • $\begingroup$ You may let $abc=1$ then let $a=\frac{x}{y},\,b=\frac{y}{z},\,c=\frac{z}{x}$ then SOS method will help you solve this inequality easy! $\endgroup$
    – NKellira
    Jun 24, 2020 at 13:47

1 Answer 1


I'll prove a stronger inequality:

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}.$$

Indeed, by C-S $$\sum_{cyc}\frac{a}{a+b}=\sum_{cyc}\frac{a^2}{a^2+ab}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}.$$ Thus, it's enough to prove that $$\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ or $$\frac{a+b+c}{(a+b+c)^2-ab-ac-bc}\geq\frac{1}{a+b+c-\sqrt[3]{abc}}$$ or $$ab+ac+bc\geq(a+b+c)\sqrt[3]{abc},$$ which says that in the last case our inequality is proven.

Now, let $$ab+ac+bc\leq(a+b+c)\sqrt[3]{abc}.$$ Thus, by C-S again we obtain: $$\sum_{cyc}\frac{a}{a+b}=\sum_{cyc}\frac{a^2c^2}{a^2c^2+c^2ab}\geq\frac{(ab+ac+bc)^2}{\sum\limits_{cyc}(a^2b^2+a^2bc)}$$ and it remains to prove that $$\frac{(ab+ac+bc)^2}{\sum\limits_{cyc}(a^2b^2+a^2bc)}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ or $$\frac{(ab+ac+bc)^2}{(ab+ac+bc)^2-(a+b+c)abc}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}},$$ which is $$ab+ac+bc\leq(a+b+c)\sqrt[3]{abc}$$ exactly.


  • 2
    $\begingroup$ +1 that's impressive $\endgroup$
    – Calvin Lin
    Dec 11, 2019 at 12:40
  • $\begingroup$ I do not understand what you claim regarding $ab+ac+bc\geq(a+b+c)\sqrt[3]{abc}$. Is this true or not? You claim the opposite in the subsequent derivation. If either inequality is true, do you not have to prove it? $\endgroup$
    – Hans
    Dec 13, 2019 at 3:47
  • $\begingroup$ @Hans I proved the starting inequality for $ab+ac+bc\geq(a+b+c)\sqrt[3]{abc}$ and I proved the starting inequality for $ab+ac+bc\leq(a+b+c)\sqrt[3]{abc}$, which says our inequality is proven! In the general $ab+ac+bc\geq(a+b+c)\sqrt[3]{abc}$ is wrong, of course. $\endgroup$ Dec 13, 2019 at 5:52
  • 1
    $\begingroup$ I see. You mean you arrive at the desired inequality in either case, or that in all cases, the desired inequality is true. Clever. +1 $\endgroup$
    – Hans
    Dec 13, 2019 at 9:43

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