Let $a,b,c>0$ satisfy $abc=1$, prove that: $$\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$$ My attempt: Let $a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}$, we have $xyz=1$ and using $abc=1$, the inequality can be written as: $$\dfrac{x}{\sqrt{x+y}}+\dfrac{y}{\sqrt{y+z}}+\dfrac{z}{\sqrt{z+x}}\le \dfrac{xy+yz+zx}{\sqrt{2}}$$ I'm trying to use Cauchy-Schwarz: $$LHS\le\sqrt{(x+y+z)(\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x})}$$ but now I have to prove $$\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}\le\dfrac{3}{2}$$ because $ab+bc+ca\ge\sqrt{3(a+b+c)}$, but I can't prove it. Can anyone give me a hint? Not necessarily a complete solution.
By the way, I also relized a problem that seems quite similar to the above problem $\sqrt{\frac{2 x}{x+y}}+\sqrt{\frac{2 y}{y+z}}+\sqrt{\frac{2 z}{z+x}} \leq 3$ if $x,y,z>0$ (Vasile Cirtoaje) (and then we can use $3\le xy+yz+zx$ ?Hope it helps)