Let $a, b, c$ be positive real number such that $a+b+c = 3$.

Prove that

$$\displaystyle\sum_{cyc}\frac{\sqrt{2}a^2b}{2a+b} \leq \displaystyle\sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$$

My attempt :


$\displaystyle\sum_{cyc}\frac{a^2b}{2a+b} \leq \displaystyle\sum_{cyc} \frac{a^2b}{3 \sqrt[3]{a^2b}} \leq \displaystyle\sum_{cyc} \frac{\sqrt[3]{a^4b^2}}{3} $ ---[1]


$a^2+2ab \geq 3\sqrt[3]{a^4b^2}$

$b^2+2bc \geq 3\sqrt[3]{b^4c^2}$

$c^2+2ca \geq 3\sqrt[3]{c^4a^2}$

$(a+b+c)^2 = 9 \geq 3 \displaystyle\sum_{cyc}\sqrt[3]{a^4b^2}$

$3 \geq \displaystyle\sum_{cyc}\sqrt[3]{a^4b^2}$

$\displaystyle\sum_{cyc} \frac{\sqrt[3]{a^4b^2}}{3} \leq 1$ ---[2]

From [1], [2], we have

$\displaystyle\sum_{cyc}\frac{\sqrt{2}a^2b}{2a+b} \leq \sqrt{2}$

Please suggest how to show that

$ \sqrt{2} \leq \displaystyle\sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$

  • $\begingroup$ from where does it come? $\endgroup$ Dec 30, 2017 at 9:48
  • $\begingroup$ @Dr. Sonnhard Graubner, source is unknown. $\endgroup$
    – user403160
    Dec 30, 2017 at 9:56

1 Answer 1


By C-S $$\sqrt2\sum_{cyc}\frac{2a^2b}{2a+b}\leq\frac{\sqrt2}{(2+1)^2}\sum_{cyc}a^2b\left(\frac{2^2}{2a}+\frac{1^2}{b}\right)=\frac{\sqrt2}{9}\sum_{cyc}(2ab+a^2)=\sqrt2.$$ Thus, it's enough to prove that $$\sum_{cyc}\frac{\sqrt{a^2+b^2}}{1+2ab}\geq\sqrt2.$$ Now, by C-S again $$\sum_{cyc}\frac{\sqrt{a^2+b^2}}{1+2ab}=\frac{1}{\sqrt2}\sum_{cyc}\frac{\sqrt{(1^2+1^2)(a^2+b^2)}}{1+2ab}\geq\frac{1}{\sqrt2}\sum_{cyc}\frac{a+b}{1+2ab}.$$ Thus, it remains to prove that $$\sum_{cyc}\frac{a+b}{1+2ab}\geq2$$ or $$\sum_{cyc}\left(\frac{(a+b)(a+b+c)}{(a+b+c)^2+18ab}-\frac{2}{9}\right)\geq0$$ or $$\sum_{cyc}\frac{7a^2+7b^2-2c^2-22ab+5ac+5bc}{1+2ab}\geq0$$ or $$\sum_{cyc}\frac{(c-a)(-7a+11b-c)-(b-c)(-7b+11a-c)}{1+2ab}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{-7b+11c-a}{1+2bc}-\frac{-7a+11c-b}{1+2ac}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)((-7b+11c-a)(1+2ac)-(-7a+11c-b)(1+2bc))}{(1+2ac)(1+2bc)}\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2(3+11c^2-ac-bc)}{(1+2ac)(1+2bc)}\geq0$$ or $$\sum_{cyc}(a-b)^2(3+11c^2-ac-bc)(1+2ab)\geq0$$ or $$\sum_{cyc}(a-b)^2(3+11c^2-c(3-c))(1+2ab)\geq0$$ or $$\sum_{cyc}(a-b)^2(4c^2-c+1)(1+2ab)\geq0.$$ Done!

  • $\begingroup$ How did you derive this line ? $$\sum_{cyc}(a-b)^2(3+11c^2-ac-bc)(1+2ab)\geq0$$ $\endgroup$
    – user403160
    Dec 30, 2017 at 11:36
  • $\begingroup$ I added something. See now. $\endgroup$ Dec 30, 2017 at 11:49
  • $\begingroup$ I got it now. Thank you for your kind help, Michael. I wish you have a good health in the coming year :) $\endgroup$
    – user403160
    Dec 30, 2017 at 13:28
  • $\begingroup$ You are welcome! Thank you! $\endgroup$ Dec 30, 2017 at 16:42

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