Let $a,b,c\ge 0: ab+bc+ca=3.$ Prove that $$\color{black}{\sqrt{\frac{a}{bc+2}}+\sqrt{\frac{b}{ca+2}}+\sqrt{\frac{c}{ab+2}}\le \sqrt{\frac{3}{2}}\cdot\sqrt{a+b+c-1}.}$$ Equality holds at $a=b=c=1.$
Context. I've created accidentally a long time ago.
Originally, I just intended to choose some constant and $(2,3,2,1)$ turns out.
Some previous solvers failed to done because a big trouble is around $a=b=70/43; c=647/6020. $
Indeed, at these specific values, we obtain $ LHS-RHS\approx -0.0004801.$
That's why I can not use C-S $\sqrt{x}+\sqrt{y}+\sqrt{z}\le \sqrt{3(x+y+z)}$ and $$\sqrt{3\left(\frac{a}{bc+2}+\frac{b}{ca+2}+\frac{c}{ab+2}\right)}\le \sqrt{\frac{3}{2}}\cdot\sqrt{a+b+c-1}$$ is not true.
Also, the following doesn't help $$\sqrt{(a+b+c)\left(\frac{1}{ab+2}+\frac{1}{bc+2}+\frac{1}{ca+2}\right)}\le \sqrt{\frac{3}{2}}\cdot\sqrt{a+b+c-1}.$$ Maybe the Bacteria method helps. We can see similar question.
Inspired by Michael Rozenberg, I've tried $$\sum_{cyc}\sqrt{\frac{a}{(bc+2)(a+k)}\cdot(a+k)}\le \sqrt{\sum_{cyc}\frac{a}{(bc+2)(a+k)}\cdot (a+b+c+3k)}.$$ Thus, we get $3k=-1 \iff k=-1/3$ which is not valid.
Can you help me to optimize this method, especially in some unsual equality cases? It would be great if the solver attach their motivated idea with answer.
I hope to see others better ideas. Also, any ideas improving the question is welcome.
Yes, Bacteria helps here!
By C-S $$\sum_{cyc}\sqrt{\frac{a}{bc+2}}\leq\sqrt{\sum_{cyc}\frac{a}{(bc+2)(5a+1)}\sum_{cyc}(5a+1)}$$ and it's enough to prove that: $$\sum_{cyc}\frac{a}{(bc+2)(5a+1)}\leq\frac{3(a+b+c-1)}{2(5(a+b+c)+3)}.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$ and $abc=w^3$.
Thus, we need to prove that: $$v^2\sum_{cyc}\frac{a}{(bc+2v^2)(5a+v)}\leq\frac{3u-2}{2(5u+v)}$$ or $f(u)\geq0,$ where $$f(u)=80u^2v^8-168uv^9-504v^{10}+60u^3v^4w^3+40u^2v^5w^3-602uv^6w^3-892v^7w^3+$$ $$+685u^2v^2w^6+349uv^3w^6-228v^4w^6-375vw^9+1125uw^9.$$ But $$f'(u)=160uv^8-168v^9+180u^2v^4w^3+80uv^5w^3-600v^6w^3+$$ $$+1370uv^2w^6+349v^3w^6+1125w^9\geq$$ $$\geq4(40uv^8-42v^9+45u^2v^4w^3+20uv^5w^3-150v^6w^3+342uv^2w^6),$$ where the last expression increases as a function of $u$ and by $uvw$ for the proof that $f'(u)\geq0$ it's enough to prove that $$40uv^8-42v^9+45u^2v^4w^3+20uv^5w^3-150v^6w^3+342uv^2w^6\geq0$$ for equality case of two variables.
Let $b=a$ and $c=\frac{3-a^2}{2a}.$
Thus, we need to prove that: $$513a^8-2565a^6-65a^5+1839a^4+130a^3+3797a^2+27a+80\geq0,$$ which is true for $0<a\leq\sqrt3$.
Id est, $f$ increases and by $uvw$ again it's enough to prove $f(u)\geq0$ for equality case of two variables, which gives $$(a-1)^2(75a^7+85a^6-276a^5-248a^4-327a^3-41a^2+1320a+420)\geq0,$$ which is true even for any non-negative value of $a$.