<s>[INST] Prove : $\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$ 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)
 [/INST] As an aside, reversing (or repeating) the change of variables further simplifies the work.
Starting off similar to OP's / River Li's work,  we WTS
$$\sqrt{ ( x+y+z) (\frac{x}{x+y} + \frac{y}{y+z} + \frac{z}{z+x}) } \leq \frac{ xy+yz+zx} { \sqrt{2} }. $$
This is equivalent to
$$ \sum x + \sum \frac{xz}{x+y} \leq \frac{ (xy+yz+zx)^2 } { 2}.$$
We revert the change of variables, letting $ x = \frac{1}{a}$ and using $abc = 1$. We WTS
$$ \sum bc + \sum \frac{ ab^2}{a+b} \leq \frac{ (a+b+c)^2}{2} \Leftrightarrow \sum \frac{ab^2}{a+b} \leq \frac{ a^2+b^2+c^2}{2}. $$
This is true because
$$ \frac{ ab^2}{a+b} \leq \frac{ b^2+ab}{4} \Rightarrow \sum \frac{ab^2}{a+b} \leq \sum \frac{b^2+ab}{4} \leq \sum \frac{ a^2}{2}.$$

Notes

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*And of course, for those who don't want to substitute twice, you can work in just $a, b, c$. However, the steps seem "less obvious".

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