If $a,b,c\ge 0: ab+bc+ca=1,$ then prove that $$\frac{1}{\sqrt{4a+bc}}+\frac{1}{\sqrt{4b+ca}}+\frac{1}{\sqrt{4c+ab}}\le \sqrt{2(a+b+c)}.$$ I saw it here.
I think the following will help. Squaring both side, it's prove $$2\sum_{cyc}\frac{1}{\sqrt{(4a+bc)(4b+ca)}}\le 2(a+b+c)-\sum_{cyc}\frac{1}{4a+bc}.$$ Now, we need to find a suitable estimate $$\sum_{cyc}\frac{1}{\sqrt{(4a+bc)(4b+ca)}}\le f(a,b,c).$$
I stopped here. If we find good enough $f(a,b,c),$ the rest is smooth by $$4f^2(a,b,c)\le \left[2(a+b+c)-\sum_{cyc}\frac{1}{4a+bc}\right]^2.$$
Hope you give me a hint to continue my approach. Also, all idea and comment are welcome.
By C-S $$\sum_{cyc}\frac{1}{\sqrt{4a+bc}}\leq\sqrt{\sum_{cyc}\frac{3a+b+c}{4a+bc}\sum_{cyc}\frac{1}{3a+b+c}}$$ and it's enough to prove that: $${\sum_{cyc}\frac{3a+b+c}{4a+bc}\sum_{cyc}\frac{1}{3a+b+c}}\leq2(a+b+c).$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v=\frac{1}{\sqrt3},$ and $abc=w^3$.
Thus, we need to prove that $f(u)\geq0,$ where $f$ increases, which by $uvw$ says that it's enough to prove $f(u)\geq0$ for equality case of two variables.
Let $b=a$ and $c=\frac{1-a^2}{2a},$ where $0<a<1$, and we obtain: $$(1-a)(21a^{10}-189a^9+199a^8-204a^7+156a^6+301a^5-87a^4-98a^3+27a^2-2a+6)\geq0,$$ which is true for $0<a<1.$
Can you get a full proof now?