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)
 A: 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

*

*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".

A: Remark: As Calvin Lin pointed out, we can just deal with $a, b, c$, without the substitutions.
We have
\begin{align*}
 &\sum_{\mathrm{cyc}}
 \sqrt{\frac{ab}{bc^2 + 1}} \\
 =\,& \sum_{\mathrm{cyc}}
 \sqrt{\frac{ab ab}{(bc^2 + 1)ab}}\\
 =\,& \sum_{\mathrm{cyc}}\frac{ab}{\sqrt{ab + bc}}\\
 \le\,& \sqrt{(ab + bc + ca)\left(\frac{ab}{ab + bc} + \frac{bc}{bc + ca} + \frac{ca}{ca + ab}\right)} \tag{1}\\[5pt]
 =\,&\sqrt{\frac{(ab + bc + ca)ab}{ab + bc} + \frac{(ab + bc + ca)bc}{bc + ca} + \frac{(ab + bc + ca)ca}{ca + ab}}\\[5pt]
 =\,& \sqrt{ab + \frac{ca^2}{a + c} + bc + \frac{ab^2}{b + a} + ca + \frac{bc^2}{c + b}}\\[5pt]
 \le\,& \sqrt{ab + \frac{\frac{(a + c)^2}{4}a}{a + c} + bc + \frac{\frac{(b + a)^2}{4}b}{b + a} + ca + \frac{\frac{(b + c)^2}{4}c}{c + b}}\tag{2}\\[5pt]
 =\,&\sqrt{\frac{1}{4}(a^2 + b^2 + c^2) + \frac54(ab + bc + ca)}
\end{align*}
where we have used the Cauchy-Bunyakovsky-Schwarz inequality in (1),
and $ca \le \frac{(c + a)^2}{4}$ etc. in (2).
It suffices to prove that
$$\frac{(a + b + c)^2}{2} \ge \frac{1}{4}(a^2 + b^2 + c^2) + \frac54(ab + bc + ca)$$
or
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
which is true.
We are done.
A: Use this inequality:
If a, b and c are all positive, show that:
$$\frac a{b+c}+\frac b{a+c}+\frac c{b+a}\geq\frac 32$$
Solution:
Since a, b and c are positive we have a+b, b+c, and c+a are also positive. Let b+c<c+a<a+b, so that $\frac 1{b+c}\geq \frac 1{c+a} \geq \frac 1{a+b}$
Using:
$(a_1+a_2+a_3+...+a_n)(b_1+b_2+b_3+...b_n)\geq n(a_1b_1+a_2b_2+a_3b_3+ ...a_nb_n)$
we have:
$[(a+b)+(b+c)+(a+c)]\cdot \big[\frac 1{ a+b} +\frac 1{b+c}+\frac 1{a+c}\big]\geq3\big[ (a+b)\cdot \frac 1 {a+b}+(b+c)\cdot \frac 1 {b+c}+(a+c)\cdot \frac 1 {a+c}\big]$
or:
$2(a+b+c)\big[\frac 1{a+b}+\frac 1{b+c} +\frac 1 {c+a}\big]\geq 9$
finally:
$\big(\frac a{b+c}+1+\frac b{c+a}+1+\frac c{a+b}+1\big)\geq \frac 9 2$
or:
$\big(\frac a{b+c}+\frac b{c+a}+\frac c{a+b}\big)\geq \frac 3 2$
Now since you let $a=\frac 1x$, $b=\frac 1y$ and $c=\frac 1z$, the conditions changes and you have to get what you wanted to prove, i. e:
$\big(\frac x{x+y}+\frac y{y+z}+\frac z{z+x}\big)\leq \frac 3 2$
