I have this :
Let $a,b,c>0$ such that $abc=1$ then we have : $$\Big(\frac{a}{a^{10}+1}+\frac{b}{b^{10}+1}+\frac{c}{c^{10}+1}\Big)\Big(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\Big)\leq \frac{9}{4}$$
I have tried to use Tchebytchev's inequality to get a sum but it doesn't work .There is a case where we have :
$$\Big(\frac{a}{a^{10}+1}+\frac{b}{b^{10}+1}+\frac{c}{c^{10}+1}\Big)\Big(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\Big)\leq\Big(\frac{a}{a^{10}+1}+\frac{b}{b^{10}+1}+\frac{c}{c^{10}+1}\Big)\frac{3}{2}\leq \frac{9}{4}$$
Furthermore I have tried Buffalo's way (full expanding and delete the constraint by deleting a variable) but It doesn't seems to work .
Finally I have tried derivative unsuccessfully.
If you have some ideas...Thanks a lot .
The BW works!
Let $a=\frac{y}{x}$ and $b=\frac{z}{y}$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and we need to prove that: $f(x,y,z)\geq0,$ where $$f(x,y,z)=\frac{9}{4}-\sum_{cyc}\frac{x^9y}{x^{10}+y^{10}}\sum_{cyc}\frac{x^2}{x^2+yz}.$$ Since $f(x,y,z)\geq0$ is a cyclic inequality we can assume that $x=\min\{x,y,z\}$.
Consider two cases.
Let $x=a$, $y=a+u$ and $z=a+u+v$, where $u$ and $v$ are non-negatives.
Thus, it's obvious that $f(a,a+u,a+u+v)\geq0$.
See here: https://www.wolframalpha.com/input/?i=9%2F4-%28x%5E9y%2F%28x%5E10%2By%5E10%29%2By%5E9z%2F%28y%5E10%2Bz%5E10%29%2Bz%5E9x%2F%28z%5E10%2Bx%5E10%29%29%28y%5E2%2F%28y%5E2%2Bxz%29%2Bz%5E2%2F%28z%5E2%2Bxy%29%2Bx%5E2%2F%28x%5E2%2Byz%29%29%2Cx%3Da%2Cy%3Da%2Bu%2Cz%3Da%2Bu%2Bv
Thus, easy to see that $f(a,a+u,a+u+v)\geq0$.
See here: https://www.wolframalpha.com/input/?i=9%2F4-%28x%5E9y%2F%28x%5E10%2By%5E10%29%2By%5E9z%2F%28y%5E10%2Bz%5E10%29%2Bz%5E9x%2F%28z%5E10%2Bx%5E10%29%29%28y%5E2%2F%28y%5E2%2Bxz%29%2Bz%5E2%2F%28z%5E2%2Bxy%29%2Bx%5E2%2F%28x%5E2%2Byz%29%29%2Cx%3Da%2Cy%3Da%2Bu%2Bv%2Cz%3Da%2Bv