Let , $a\geq b \geq c >0$ such that $a+b+c=\frac{10}{11}$ then prove (or disprove) :
$$\frac{\cosh(a)}{b+c}+\frac{\sinh(b)}{a+c}+\frac{\tanh(c)}{a+b}>e$$
We cannot use Jensen's inequality as proof for this inequality since the functions $\cosh,\sinh,\tanh$ are differents .
The use of Am-Gm is inefficient . Moreover I have tried something like :
$$\frac{\cosh(a)}{b+c}+\frac{\sinh(b)}{a+c}+\frac{\tanh(c)}{a+b}>\frac{2e}{3}\Big(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\Big)\geq e$$
Without success .
As a last resort I have tried power series but it's very ugly so not nice .
Update : 03/11/2020
I simplify a little bit the proof of user RiverLi wich is a nice proof .
Due to convexity ($\cosh$ and $\sinh$) or concavity ($\tanh$) we have the following inequalities (always with the constraints above) :
$$\frac{\cosh(a)}{\frac{10}{11}-a}\geq f(a)=\frac{\cosh(\frac{10}{33})+(a-\frac{10}{33})\sinh(\frac{10}{33})}{\frac{10}{11}-a}$$
$$\frac{\sinh(b)}{\frac{10}{11}-b}\geq g(b)=\frac{\sinh(\frac{10}{33})+(b-\frac{10}{33})\cosh(\frac{10}{33})}{\frac{10}{11}-b}$$
$$\frac{\tanh(c)}{\frac{10}{11}-c}\geq h(c)=\frac{c\tanh(\frac{10}{33})\frac{33}{10}}{\frac{10}{11}-c}$$
Now it's less harder to show that the function $f,g,h$ are convex and continue the proof of RiverLi .
Any idea is welcome!
Thanks in advance
Hint.
Let $f(a) = \frac{\cosh a}{10/11 - a}$, $g(b) = \frac{\sinh b}{10/11 - b}$ and $h(c) = \frac{\tanh c}{10/11 - c}$.
All $f, g, h$ are convex functions.
Then we have $f(a) \ge f(10/33) + f'(10/33)(a - 10/33)$, $g(b) \ge g(10/33) + g'(10/33)(b-10/33)$ and $h(c) \ge h(10/33) + h'(10/33)(c-10/33)$.
It suffices to prove that $$f(10/33) + f'(10/33)(a - 10/33) + g(10/33) + g'(10/33)(b-10/33) + h(10/33) + h'(10/33)(c-10/33) > \mathrm{e}.$$ Since $a\ge b\ge c$ and $f'(10/33) > g'(10/33) > h'(10/33)$, we have $$\mathrm{LHS} \ge f(10/33) + g(10/33) + h(10/33) > \mathrm{e}.$$