Prove $\frac{1}{\sqrt{2a+b+c}}+\frac{1}{\sqrt{2b+a+c}}+\frac{1}{\sqrt{2c+b+a}}\le \frac{9}{2}\cdot\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$

Question

Let $a,b,c>0.$ Prove that $$\frac{1}{\sqrt{2a+b+c}}+\frac{1}{\sqrt{2b+a+c}}+\frac{1}{\sqrt{2c+b+a}}\le \frac{9}{2}\cdot\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$$ WLOG, assuming that $a+b+c=1$ and we'll prove $$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\cdot \left(\frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}+\frac{1}{\sqrt{c+1}}\right)\le \frac{9}{2}.$$ By using Cauchy-Schwarz inequality $$\frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}+\frac{1}{\sqrt{c+1}}\le \sqrt{3\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)},$$so we need to prove $$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\sqrt{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}}\le \frac{3\sqrt{3}}{2},$$ or $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le \frac{27}{4}\cdot \frac{1}{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}.$$ I was stuck here. Hope you help me continue my approach.

Any ideas and comments are welcome.

Answer 1

Proof.

By using Cauchy-Schwarz $$\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a+1}}=\frac{\sqrt{3}\cdot\left(\frac{1}{\sqrt{3}}\cdot\sqrt{a}+\sqrt{b}\cdot\frac{1}{\sqrt{3}}+\sqrt{c}\cdot\frac{1}{\sqrt{3}}\right)}{\sqrt{a+1}}$$ $$\le \sqrt{\frac{3}{a+1}\cdot\left(\frac{1}{3}+b+c\right)\cdot\left(\frac{2}{3}+a\right)}=\sqrt{-3a-\frac{7}{3(a+1)}+5}.$$ Similarly, $$\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{b+1}}\le \sqrt{-3b-\frac{7}{3(b+1)}+5}.$$$$\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{c+1}}\le \sqrt{-3c-\frac{7}{3(c+1)}+5}.$$ Thus, it's enough to prove $$\sum_{cyc}\sqrt{-3a-\frac{7}{3(a+1)}+5}\le \frac{9}{2}. \tag{*}$$ Also by Cauchy-Schwarz $$\sum_{cyc}\sqrt{-3a-\frac{7}{3(a+1)}+5}\le \sqrt{3\left(-3(a+b+c)+15-\frac{7}{3}\cdot\sum_{cyc}\frac{1}{a+1}\right)},$$and $$\frac{1}{a+1}\ge \frac{9}{a+b+c+3}=\frac{9}{4}.$$ Hence, the $(*)$ is proven and we complete the proof here.

Answer 2

The Bacteria helps.

We need to prove that: $$\sum_{cyc}\frac{1}{\sqrt{2a^2+b^2+c^2}}\leq\frac{9}{2(a+b+c)},$$ where $a$, $b$ and $c$ are positives.

Indeed, by C-S $$\sum_{cyc}\frac{1}{\sqrt{2a^2+b^2+c^2}}\leq\sqrt{\sum_{cyc}\frac{1}{(2a^2+b^2+c^2)(a+2b+2c)}\sum_{cyc}(a+2b+2c)}$$ and it's enough to prove that: $$\sum_{cyc}\frac{1}{(2a^2+b^2+c^2)(a+2b+2c)}\leq\frac{81}{20(a+b+c)^3}$$ and the rest is smooth.

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that: $$\frac{\sum\limits_{cyc}(9u^2-6v^2+a^2)(9u^2-6v^2+b^2)(6u-a)(6u-b)}{\prod\limits_{cyc}(9u^2-6v^2+a^2)\prod\limits_{cyc}(6u-a)}\leq\frac{3}{20u^3}$$ or $$\tfrac{(-1080u^3+900uv^2)w^3+A(u,v^2)}{\left(w^6-(54u^3-36uv^2)w^3+1456u^6-2916u^4v^2+2025u^2v^4-486v^6\right)(108u^3+18uv^2-w^3)}\leq\frac{3}{20u^3}$$ or $f(w^3)\geq0,$ where $$f(w^3)=3\left(w^6-(54u^3-36uv^2)w^3+1456u^6-2916u^4v^2+2025u^2v^4-486v^6\right)(108u^3+18uv^2-w^3)-20u^3((-1080u^3+900uv^2)w^3+A(u,v^2)),$$ where $A$ is a some polynomial of $u$ and $v^2$.

But, $$f'(w^3)=3(2w^3-54u^3+36uv^2)(108u^3+18uv^2-w^3)+$$$$+3(w^6-(54u^3-36uv^2)w^3+1456u^6-2916u^4v^2+2025u^2v^4-486v^6)\cdot(-1)+$$ $$+20u^3(1080u^3-900uv^2)=$$ $$=-3(88u^6+168u^4v^2+1377u^2v^4+1377u^2v^4-486v^6-324u^3w^3+36uv^2w^3+3w^6)<0,$$ which says that $f$ decreases.

Thus, by $uvw$ it's enough to prove $f(w^3)\geq0$ for equality case of two variables.

Since it's homogeneous and for $b=c\rightarrow0^+$ it's obvious, it's enough to assume $b=c=1$, which gives $$\frac{2}{(a^2+3)(2a+3)}+\frac{1}{(2a^2+2)(a+4)}\leq\frac{81}{20(a+2)^3}$$ or $$(a-1)^2(102a^4+545a^3+648a^2+705a+916)\geq0$$ and we are done.

Answer 3

Another way(iura).

Let $f(x)=3\sqrt{x}+\frac{2}{\sqrt{1+x}},$ where $x\in(0,1).$

Thus, $f$ is a concave function and by Jensen we obtain: $$3(\sqrt{a}+\sqrt{b}+\sqrt{c})+2\sum_{cyc}\frac{1}{\sqrt{1+a}}\leq6\sqrt3$$ and after using AM-GM we are done!