how to prove this inequality with $a,b,c\in [1,3]$ let $a,b,c\in [1,3]$,show that
$$3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{45}{a+b+c}\ge 16\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)$$
I had found this simaler problem,https://artofproblemsolving.com/community/c6h615194
I can't prove this it

WLOG, assume that $c = \max(a,b,c) = 3$. The inequality becomes
$$\frac{3}{a} - \frac{16}{a+3} + \frac{3}{b} - \frac{16}{b+3} + 1 - \frac{16}{a+b} + \frac{45}{a+b+3} \ge 0.$$
Let $f(x) = \frac{3}{x} - \frac{16}{x+3}$.
We have $f''(x) = \frac{6}{x^3} - \frac{32}{(x+3)^3}$.
For $x\in [1,3]$, we have $(x+3)^3/x^3 = (1 + 3/x)^3 \ge 2^3 > 32/6$
and hence $f''(x) > 0$.
Thus, $f(x)$ is convex on $[1,3]$. 
We have
$f(a) + f(b) \ge 2 f(\frac{a+b}{2})$. It suffices to prove that
$$\frac{12}{a+b} - \frac{64}{a+b+6}  + 1 - \frac{16}{a+b} + \frac{45}{a+b+3} \ge 0$$
or
$$\frac{(a+b-2)(a+b-6)^2}{(a+b)(a+b+3)(a+b+6)}\ge 0.$$
We are done.
