a,b,c>0 and prove $\frac{a}{a+\sqrt{(a+2b)(a+2c)}}+\frac{b}{b+\sqrt{(b+2a)(b+2c)}}+\frac{c}{c+\sqrt{(c+2b)(c+2a)}}\le \frac{3}{4}.$

Question

Let $a,b,c>0$. Prove that $$\frac{a}{a+\sqrt{(a+2b)(a+2c)}}+\frac{b}{b+\sqrt{(b+2a)(b+2c)}}+\frac{c}{c+\sqrt{(c+2b)(c+2a)}}\le \frac{3}{4}.$$ It is from a book. My tries did not lead to anything ineteresting thing.

I tried Cauchy- Schwarz as $\sqrt{(a+b+b)(c+a+c)}\ge \sqrt{ab}+\sqrt{cb}+\sqrt{ac},$ and we need to prove $$\sum_{cyc}\frac{a}{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})}\le \frac{3}{4}.$$It is equivalent to $\sum{cyc}\sqrt{ab}(\sqrt{a}+\sqrt{b})\le 6\sqrt{abc},$which is obviously wrong by AM-GM.

Also I used AM-HM: $\sqrt{(a+2b)(a+2c)}\ge \dfrac{(a+2b)(a+2c)}{a+b+c}$ without success.

Hope to see some good ideas. Thank you.

Answer 1

We need to prove that: $$\sum_{cyc}\frac{a\left(\sqrt{(a+2b)(a+2c)}-a\right)}{2ab+2ac+4bc}\leq\frac{3}{4}$$ or $$\sum_{cyc}\frac{a\sqrt{(a+2b)(a+2c)}}{ab+ac+2bc}\leq\frac{3}{2}+\sum_{cyc}\frac{a^2}{ab+ac+2bc}.$$ Now, by C-S $$\sum_{cyc}\frac{a\sqrt{(a+2b)(a+2c)}}{ab+ac+2bc}\leq\sqrt{\sum_{cyc}\frac{a}{ab+ac+2bc}\sum_{cyc}\frac{a(a+2b)(a+2c)}{ab+ac+2bc}}$$ and it's enough to prove that: $$\left(\frac{3}{2}+\sum_{cyc}\frac{a^2}{ab+ac+2bc}\right)^2\geq\sum_{cyc}\frac{a}{ab+ac+2bc}\sum_{cyc}\frac{a(a+2b)(a+2c)}{ab+ac+2bc}$$ or $$\frac{9}{4}+3\sum_{cyc}\frac{a^2}{ab+ac+2bc}+\left(\sum_{cyc}\frac{a^2}{ab+ac+2bc}\right)^2\geq$$ $$\geq\sum_{cyc}\frac{a}{ab+ac+2bc}\sum_{cyc}\left(\frac{a^3}{ab+ac+2bc}+2a\right)$$ or $$\tfrac{9}{4}+\sum_{cyc}\tfrac{3a^2-2a^2-2ab-2ac}{ab+ac+2bc}\geq\sum_{cyc}\tfrac{a}{ab+ac+2bc}\sum_{cyc}\tfrac{a^3}{ab+ac+2bc}-\left(\sum_{cyc}\tfrac{a^2}{ab+ac+2bc}\right)^2$$ or $$\frac{9}{4}+\sum_{cyc}\frac{a^2-2ab-2ac}{ab+ac+2bc}\geq\sum_{cyc}\frac{a^3b+b^3a-2a^2b^2}{(ab+ac+2bc)(ab+bc+2ac)}$$ or $$\sum_{sym}(a^3b^3+6a^4bc-9a^3b^2c+2a^2b^2c^2)\geq0,$$ which is true by Schur and Muirhead.

Answer 2

Proof.

We'll prove the following isolated fudging $$\frac{a}{a+\sqrt{(a+2b)(a+2c)}}\le \frac{a}{4(a+b)}+\frac{a}{4(a+c)}.\tag{*}$$

Take cyclic sum on $(*),$ we get desired inequality.

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Indeed, it's $$\frac{1}{a+\sqrt{(a+2b)(a+2c)}}\le \frac{1}{4(a+b)}+\frac{1}{4(a+c)}.$$

or

$$(2a+b+c)\left(a+\sqrt{(a+2b)(a+2c)}\right)\ge 4(a+b)(a+c),$$or $$\left(\frac{a+b}{a}+\frac{a+c}{a}\right)\left(1+\sqrt{\frac{a+2b}{a}\cdot\frac{a+2c}{a}}\right)\ge 4\cdot\frac{a+b}{a}\cdot\frac{a+c}{a}.$$ Let $x=\dfrac{a+b}{a}>0;y=\dfrac{a+c}{a}>0$ then $x+y>2.$

We rewrite the inequality as $$(x+y)[1+\sqrt{(2x-1)(2y-1)}]\ge 4xy$$ $$\iff x+y-\frac{4xy}{x+y}\ge x+y-1-\sqrt{(2x-1)(2y-1)} $$ or $$(x-y)^2\left[\frac{1}{x+y}-\frac{1}{x+y+\sqrt{(2x-1)(2y-1)}-1}\right]\ge 0.$$ Thus, it's enough to prove $$\sqrt{(2x-1)(2y-1)}-1\ge 0 \iff (2x-1)(2y-1)\ge 1$$ $$\iff 2xy\ge x+y\iff ab+ac+2bc\ge 0,$$which is obvious.