<s>[INST] Inequality $\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}+\frac{b}{\sqrt{ca}}\cdot\frac1{b+1}+\frac{c}{\sqrt{ab}}\cdot\frac1{c+1}\leqslant\sqrt2.$ Let $a,b,c>0$ and $\frac1{a+1}+\frac1{b+1}+\frac1{c+1}=1.\qquad $ Prove $$\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}+\frac{b}{\sqrt{ca}}\cdot\frac1{b+1}+\frac{c}{\sqrt{ab}}\cdot\frac1{c+1}\leqslant\sqrt2.$$
The full solution (due to @Hanno) see at the very end in $\underline{\textbf{Conclusion}}\,\,$ 
My attemps:


*

*AM>HM gives $a+b+c\geqslant 6.$ Put $a=a_1t,b=b_1t,c=c_1t, \,t\geqslant 1, \, a_1+b_1+c_1=6$


As $\frac 1{x+1}$ is decreasing function we have $\sum\limits_{cyc}\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}\leqslant \sum\limits_{cyc}\frac{a_1}{\sqrt{b_1c_1}}\cdot\frac1{a_1+1}.$
So is it possible to prove the same inequality with new restriction $a+b+c=6?$


*$a=\tan^2\alpha, b=\tan^2\beta, c=\tan^2\gamma$ gives 


$$\sum\limits_{cyc}\sin^2\alpha\cot\beta\cot\gamma\leqslant \sqrt2$$ 
if $0<\alpha,\beta,\gamma<\frac\pi2$ and $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1.$


*After hint of Michael Rozenberg. Rewrite the substitution as linear system with determinant $d=\begin{vmatrix}-a&1&1\\ 
1&-b&1\\ 1&1&-c\\ \end{vmatrix}$


If $d\ne 0$ then $x=y=z=0$ and so the hint doesn't work
The equality $d=0$ gives editional restriction, but I am not sure that this case deserves attention.
In fact, I have an ugly solution with brute force, but I want something beautiful. 


*Continue. Agree with $a=\frac{y+z}{x}$, $b$ and $c$ cyclically and $x+y+z=1.$ I'll try not to break the symmetry as in my ugly solution.


Substution gives
 $$\sum\limits_{cyc}\frac{(y+z)\sqrt{yz}}{\sqrt{(x+z)(x+y)}}\leqslant\sqrt2$$
squaring gives
$$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+2\frac{(z+x)\sqrt{zx}}{\sqrt{(y+x)(y+z)}}\cdot\frac{(x+y)\sqrt{xy}}{\sqrt{(z+y)(z+x)}}\right)\leqslant2$$
or
$$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+
2\frac{x\sqrt{yz(x+y)(x+z)}}{z+y}\right)\leqslant2$$
I do not see here where to use AM-GM, or did I miss the moment?
Thanks,  @Michael Rozenberg, very cool.  I did not miss, I did not get to the point, although I was already near. Why I stubbornly did not want to multiply by the common denominator? Although further my restriction $x+y+z=1$ would not help but would only hinders.


*Yesterday I did not check the last inequality but about this later.


We  conside the case $d=0.$  In order for the linear system to have a unique solution, it is necessary to add a new constraint. I put $x+y+z=1$ and violated homogenity. To save it let $x+y+z=I$ where $I$ denotes the sum only. 
Then from the system we have $x=\frac I{a+1}$ and $a=\frac Ix-1=\frac{y+z}{x}$ i.e. we simply forget that $I=1$ and save the homogenity.
we amend the previous one:
$$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+
2\frac{x\sqrt{yz(x+y)(x+z)}}{z+y}\right)\leqslant2I^2$$
After AM-GM we have
$$\sum\limits_{cyc}\left({yz(y+z)^3}+
x(x+y)(x+z)(xy+xz+2yz)\right)\leqslant2I^2\prod\limits_{cyc}(x+y)$$
It remains to check the last inequality.
For this its sufficent to calculete the all summands in both sides. I do not want to apply brutal forse.
Avidently both sides have can have members only of type $x^\alpha y^\beta z^\gamma$ with $\alpha+\beta+\gamma=5$ and by symmetry we may assume $\alpha\leqslant\beta\leqslant\gamma.$ My quation is: Is there another way?
Start with memmber $yz^4$. It is contained in the LHS but not contained in the right. So the  last inequality must contain negative term and the proof must be corrected if it is possibble yet. Or I'm wrong. 
$\underline{\textbf{Conclusion}}\,\,$ The inequality turned out to be interesting and the proof  gave @Hanno. Allow yourself to make some minor corrections.
The original restriction was
$$\sum_{\text{cyc}}{1\over a+1} =1 \qquad (1)$$
Now $\sum_{\text{cyc}}{a\over a+1} =\sum_{\text{cyc}}{a+1-1\over a+1} =3-1=2.$ So 
$$\sum_{\text{cyc}}{a\over a+1} =2 \qquad (2)$$
Apply the Cauchy–Bunyakovsky–Schwarz inequality (CBS):
$$\sum_{\text{cyc}}{a\over\sqrt{bc}}\cdot\frac1{a+1}
\:= \:
\sum_{\text{cyc}}\sqrt{{a\over a+1}}\cdot\sqrt{\frac a{(a+1)bc}}\:\leqslant\:
{\sqrt{\sum_{\text{cyc}}{a\over a+1}}}\,\cdot\,
\sqrt{\sum_{\text{cyc}}{a\over (a+1)bc}}$$
Taking into account (2) it remains to proof inequality $\sum\limits_{\text{cyc}}{a\over (a+1)bc}\:{\leqslant}\:1$ which is equivalent to
$$\sum_{\text{cyc}}{a^2\over a+1}\:\leqslant\: abc $$
We have $\sum\limits_{\text{cyc}}{a^2\over a+1}=\sum\limits_{\text{cyc}}{a^2-1+1\over a+1}=\sum\limits_{\text{cyc}}(a-1)+\sum\limits_{\text{cyc}}{1\over a+1}=a+b+c-2<a+b+c+2=abc. $
The last equality $a+b+c+2=abc$ follows from (1) by bringing to a common denominator. DONE!
The inequality is exact. Take small $a$ and $b=c=1+\frac2a.$ Then (1) is true and the LHS of required inequality evaluates to  ${\sqrt{a+2}\over a+1}\:+\:{a^2\over (a+1)(a+2)}$ which tends to $\sqrt2$ if $a\to0$.
 [/INST] The hint.
After substitution $a=\frac{y+z}{x}$, $b=\frac{x+z}{y}$ and $c=\frac{x+y}{z}$, where $x$, $y$ and $z$ are positives,
and squaring of the both sides use AM-GM.
Indeed, let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y}$, where $x$, $y$ and $z$ are positives.
Thus, the condition gives $$\frac{1}{\frac{y+z}{x}+1}+\frac{1}{\frac{x+z}{y}+1}+\frac{1}{c+1}=1$$ or
$$\frac{x+y}{x+y+z}+\frac{1}{c+1}=1$$ or
$$\frac{1}{c+1}=\frac{z}{x+y+z}$$ or
$$\frac{1}{c+1}=\frac{1}{\frac{x+y}{z}+1},$$ which gives $c=\frac{x+y}{z}$ and we need to prove that
$$\sum_{cyc}\frac{\frac{y+z}{x}}{\sqrt{\frac{(x+y)(x+z)}{yz}}\left(\frac{y+z}{x}+1\right)}\leq\sqrt2$$ or
$$\sum_{cyc}\frac{(y+z)\sqrt{yz}}{\sqrt{(x+y)(x+z)}}\leq\sqrt2(x+y+z)$$ or
$$\sum_{cyc}\sqrt{yz(y+z)^3}\leq(x+y+z)\sqrt{2(x+y)(x+z)(y+z)}$$ or
$$\sum_{cyc}\left(xy(x+y)^3+2x(x+y)(x+z)\sqrt{(x+y)(x+z)yz}\right)\leq2(x+y+z)^2\prod_{cyc}(x+y)$$ and since by AM-GM
$$2\sqrt{(x+y)(x+z)yz}=2\sqrt{(xz+yz)(xy+yz)}\leq xz+yz+xy+yz=xy+xz+2yz,$$ it's enough to prove that
$$\sum_{cyc}\left(xy(x+y)^3+x(x+y)(x+z)(xy+xz+2yz)\right)\leq2(x+y+z)^2\prod_{cyc}(x+y),$$
which is $$\sum_{cyc}(x^3y^2+x^3z^2+4x^3yz+6x^2y^2z)\geq0.$$
Done!
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