2
$\begingroup$

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 :

By AM-GM,

$\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]

By AM-GM,

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

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

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

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

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

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

$\endgroup$
  • $\begingroup$ from where does it come? $\endgroup$ – Dr. Sonnhard Graubner Dec 30 '17 at 9:48
  • $\begingroup$ @Dr. Sonnhard Graubner, source is unknown. $\endgroup$ – carat Dec 30 '17 at 9:56
1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ How did you derive this line ? $$\sum_{cyc}(a-b)^2(3+11c^2-ac-bc)(1+2ab)\geq0$$ $\endgroup$ – carat Dec 30 '17 at 11:36
  • $\begingroup$ I added something. See now. $\endgroup$ – Michael Rozenberg Dec 30 '17 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 :) $\endgroup$ – carat Dec 30 '17 at 13:28
  • $\begingroup$ You are welcome! Thank you! $\endgroup$ – Michael Rozenberg Dec 30 '17 at 16:42

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.