Prove $\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$

Question

For $a,\,b,\,c>0$. Prove: $$\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$$
My work:
After a lot of caculates, I found:

$\text{RHS-LHS}=$

However, it's hard to find in a competition.

So I wanna to find a simple way for it without Buffalo Way! Thanks a lot!

Answer 1

By C-S $$\sum_{cyc}\left(\frac{a}{b}-\frac{a}{b+c}\right)=\sum_{cyc}\frac{ac}{b(b+c)}=\frac{1}{abc}\sum_{cyc}\frac{a^2c^2}{b+c}\geq$$ $$\geq\frac{(ab+ca+bc)^2}{2abc(a+b+c)}\geq\frac{3}{2}.$$