$a,b,c>0$, prove: $\sum_{cyc}{\frac{a+\sqrt{bc}}{\sqrt{(a+b)(a+c)}+\sqrt{bc}}}\le2$

Question

Let $a,b,c>0$. Prove that: $$\!\!\frac{a+\sqrt{bc}}{\sqrt{(a+b)(a+c)}+\sqrt{bc}}+ \frac{b+\sqrt{ca}}{\sqrt{(b+c)(b+a)}+\sqrt{ca}}+ \frac{c+\sqrt{ab}}{\sqrt{(c+a)(c+b)}+\sqrt{ab}}\le2$$

My approach using AM-GM: $\sqrt{(a+b)(a+c)}\ge2\sqrt[4]{a^2bc}$ ; so we need to prove that: $$\sum_{cyc}{\frac{a+\sqrt{bc}}{2\sqrt[4]{a^2bc}+\sqrt{bc}}}\le2$$ Due to homogenious, I denote $abc=1$ which implies the new one variable inequality: $$\sum_{cyc}{\frac{a+\dfrac{1}{\sqrt{a}}}{2\sqrt[4]{a}+\dfrac{1}{\sqrt{a}}}}\le2$$ I am trying to find suitable term to finish my idea.

Is there any good way to full of my approach or other better idea? Thanks for help.

Remark. We have two ineqialities $$\sum_{cyc}\frac{a}{a+\sqrt{(a+b)(a+c)}}\leq 1$$ $$\sum_{cyc}\frac{\sqrt{bc}}{a+\sqrt{(a+b)(a+c)}}\geq 1$$ Is there a relation here ?

Answer 1

Proof.

We'll prove an isolated fudging$$\frac{a+\sqrt{bc}}{\sqrt{(a+b)(a+c)}+\sqrt{bc}}\le \frac{2a+b+c}{2(a+b+c)}.\tag{1}$$ By AM-GM inequality, \begin{align*} &\frac{a+\sqrt{bc}}{\sqrt{(a+b)(a+c)}+\sqrt{bc}}\\&=\frac{(a+\sqrt{bc})(\sqrt{(a+b)(a+c)}-\sqrt{bc})}{a(a+b+c)}\\&=\frac{2(a+\sqrt{bc})\sqrt{(a+b)(a+c)}-2\sqrt{bc}(a+\sqrt{bc})}{2a(a+b+c)}\\&\le \frac{(a+\sqrt{bc})^2+(a^2+ab+bc+ca)-2\sqrt{bc}(a+\sqrt{bc})}{2a(a+b+c)}\\&=\frac{2a+b+c}{2(a+b+c)}. \end{align*} Take cyclic sum on $(1)$ we obtain the desired result. Equality holds iff $a=b=c>0.$

Answer 2

Alternative proof.

We may use AM-GM as $$2\sqrt{(a+b)(a+c)}=2\sqrt{\frac{a^2+ab+bc+ca}{a+\sqrt{bc}}(a+\sqrt{bc})}\le \frac{a^2+ab+bc+ca}{a+\sqrt{bc}}+a+\sqrt{bc}$$ $$=\frac{a^2+ab+bc+ca}{a+\sqrt{bc}}+a-\sqrt{bc}+2\sqrt{bc}=\frac{a(2a+b+c)}{a+\sqrt{bc}}+2\sqrt{bc}.$$ It implies $$\frac{a(2a+b+c)}{a+\sqrt{bc}}\ge 2\left(\sqrt{(a+b)(a+c)}-\sqrt{bc}\right)=\frac{2(a+b+c)}{\sqrt{(a+b)(a+c)}+\sqrt{bc}}.$$ Thus, $$\sum_{cyc}\frac{a+\sqrt{bc}}{\sqrt{(a+b)(a+c)}+\sqrt{bc}}\le \sum_{cyc}\frac{2a+b+c}{2(a+b+c)}=2$$and we are done!