Problem. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that$$\color{black}{a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\le \sqrt{a+b+c+2}. }$$
Source: JK.
I checked that equality holds at $abc=0.$
Indeed, when $abc=0:$ it's obviously true $b\sqrt{c}+c\sqrt{b}=\sqrt{b}+\sqrt{c}=\sqrt{b+c+2}.$
I've tried to use CBS inequality without success.
I used as $$\sum_{cyc}a\sqrt{b+c}=\sum_{cyc}\sqrt{a}.\sqrt{ab+ac}\le \sqrt{2(a+b+c)}\le \sqrt{a+b+c+2},$$which implies that the original inequality is true when $\sqrt{3}\le a+b+c\le 2.$
We need to prove in remain case $a+b+c\ge 2.$ I've just done it so far.
How can I continue my idea? Thanks for your help.
Best Answer
Proof. We use the so-called isolated fudging.
The desired inequality is written as $$\sum_{\mathrm{cyc}} a\sqrt{\frac{b + c}{a + b + c + 2}} \le 1$$ or $$\sum_{\mathrm{cyc}} \left(a\sqrt{\frac{b + c}{a + b + c + 2}} - \frac{a + a(b + c)}{a + b + c + 2}\right) \le 0 \tag{1}$$ where we use $\sum_{\mathrm{cyc}} \frac{a + a(b + c)}{a + b + c + 2} = \frac{a + b + c + 2(ab + bc + ca)}{a + b + c + 2} = 1$.
Note that $$\left(\frac{a + a(b + c)}{a + b + c + 2}\right)^2 - a^2\cdot \frac{b + c}{a + b + c + 2} = \frac{a^2(1 - ab - ac)}{(a + b + c + 2)^2} \ge 0.$$ Thus, (1) is true.
We are done.