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
 A: 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.
A: You haven't given details of what you have tried, so I shall just provide a suggestion.
One thing that will help is that if you replace $(a, b, c)$ with $(ka, kb, kc)$ for some $k > 0$ then the desired inequality is unchanged (the $k$ cancels). So given any triple $(a, b, c)$ you can scale it down so that the smallest element is equal to $1$. However, the inequality is also invariant under permutations of $(a, b, c)$. So without loss of generality you can assume that $c = 1$.
You now have an inequality in two variables. Change this into an "$f(a,b) \ge 0$" statement. At the very least, you can simply plot this in a two-dimensional plane.
A possible help is that if we write $\alpha := 1/a \in [\tfrac13, 1]$ then
$$
\frac1{a+x} = \frac1{a(1+x/a)} = \frac1a \cdot \frac1{1+x/a} = \frac{\alpha}{1+\alpha x}.$$
This last reducing may or may not be useful.
A: $$3\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)+\frac{45}{a+b+c} ≥ 16\big(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\big)$$
Let:
$\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)=A$
$\frac{9}{a+b+c}=B$
$2\big(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\big)=C$
We use inequality:
$$C=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}≥\frac{9}{2(a+b+c)}=\frac{B}{2}$$
So we must show:
$3A+5B ≥ 16 C≥ 8B$, or $3A +5B≥ 8B$, or $A≥B$ 
Or: 
$\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)≥\frac{9}{a+b+c}$
Which is easy to prove.
Note:If you want the proof of inequality used, ask it as a question, I will answer.
