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

$$3\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)+\frac{45}{a+b+c} ≥ 16\big(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\big)$$
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.
