If $ab+bc+ca=3,$ prove $\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{c+b}}+\frac{1}{\sqrt{a+c}}\le \sqrt{\frac{2(a+b+c)+21}{6}}.$

Question

If $a,b,c\ge 0: ab+bc+ca=3,$ then prove $$\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{c+b}}+\frac{1}{\sqrt{a+c}}\le \sqrt{\frac{2(a+b+c)+21}{6}}.$$ I've tried to use Cauchy-Schwarz and it's $$\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{c+b}\le\frac{2(a+b+c)+21}{18}$$but it is wrong when $a=b\rightarrow 0;c\rightarrow +\infty.$

Is there any better ideas? Hope you share to us. Thank you.

Answer 1

By C-S $$\sum_{cyc}\frac{1}{\sqrt{a+b}}=\sqrt{\sum_{cyc}\left(\frac{1}{a+b}+\frac{2}{\sqrt{a^2+3}}\right)}=$$ $$==\sqrt{\sum_{cyc}\left(\frac{1}{a+b}+\frac{4}{\sqrt{(1+3)(a^2+3)}}\right)}\leq\sqrt{\sum_{cyc}\left(\frac{1}{a+b}+\frac{4}{a+3}\right)}$$ and it's enough to prove that: $$\sum_{cyc}\left(\frac{1}{a+b}+\frac{4}{a+3}\right)\leq\frac{2(a+b+c)+21}{6}.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, since $$\sum_{cyc}\left(\frac{1}{a+b}+\frac{4}{a+3}\right)=\frac{\sum\limits_{cyc}(a+b)(a+c)}{\prod\limits_{cyc}(a+b)}+\frac{4\sum\limits_{cyc}(a+3)(b+3)}{\prod\limits_{cyc}(a+3)}=$$ $$=\frac{a^2+b^2+c^2+12}{9uv^2-w^3}+\frac{4(6(a+b+c)+30}{w^3+9v^2+27u+27},$$ we see that we need to prove $f(w^3)\geq0,$ where $f$ is a concave function.

Id est, by $uvw$ it's enough to check two cases:

1. $abc=0$;

2. Two variables are equal.

Can you end it now?

Answer 2

Some thoughts.

Denote $x=\sqrt{ab};y=\sqrt{cb};z=\sqrt{ac}\implies x,y,z>0: x^2+y^2+z^2=3.$

The original inequality becomes $$\color{black}{\sum_{cyc}\frac{1}{\sqrt{\frac{xz}{y}+\frac{xy}{z}}}\le \sqrt{\frac{2(\frac{xz}{y}+\frac{xy}{z}+\frac{yz}{x})+21}{6}}, }$$ or $$\color{black}{\sum_{cyc}\sqrt{\frac{yz}{x(y^2+z^2)}}\le \sqrt{\frac{2\dfrac{(xz)^2+(xy)^2+(yz)^2}{xyz}+21}{6}}, }$$ or $$\color{black}{\sum_{cyc}\frac{yz}{\sqrt{y^2+z^2}}\le \sqrt{\frac{2\left[(xz)^2+(xy)^2+(yz)^2\right]+21xyz}{6}}. }$$ We'll prove it's true and equality holds at $x=y=z=1$ and $x=y=\sqrt{\dfrac{3}{2}};z=0.$