How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) \ge 3/2$ Suppose $a>0, b>0, c>0$. 
Prove that: 
$$a+b+c \ge \frac{3}{2}\cdot [(a+b)(a+c)(b+c)]^{\frac{1}{3}}$$

Hence or otherwise prove: 
  $$\color{blue}{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}}$$

 A: Hint: Substitute $$b+c=x,a+c=y,a+b=z$$ so $$a=\frac{-x+y+z}{2}$$
$$b=\frac{x-y+z}{2}$$
$$c=\frac{x+y-z}{2}$$
And we get
$$\frac{-x+y+z}{2x}+\frac{x-y+z}{2y}+\frac{x+y-z}{2z}\geq \frac{3}{2}$$
Can you finish?
And we get $$\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}\geq 6$$
A: Using AM-GM inequality:$$\frac{x_1+\dots+x_n}{n}\geq (x_1\dots x_n)^{1/n}$$
let $n=3$ and $x_1=a+b,\; x_2=b+c,\; x_3=c+a$:
$$\frac{2(a+b+c)}{3}\geq [(a+b)(b+c)(c+a)]^{1/3}$$
A: Hint: 
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}\iff \frac{a+b+c}{b+c}+\frac{b+a+c}{a+c}+\frac{c+a+b}{a+b}\ge \frac{3}{2}+3$$
$$\iff \big(a+b+c\big)\cdot \bigg(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\bigg)\ge \frac{9}{2}\iff \color{blue}{\frac{2\cdot (a+b+c)}{3}\ge \frac{3}{\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}}}$$ Which is trivial by the AM-HM inequality. Done!
A: To minimize $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ , we need to have
$$
\begin{align}
0
&=\delta\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\\
&=\left(\frac1{b+c}-\frac{b}{(c+a)^2}-\frac{c}{(a+b)^2}\right)\delta a\\
&+\left(\frac1{c+a}-\frac{c}{(a+b)^2}-\frac{a}{(b+c)^2}\right)\delta b\\
&+\left(\frac1{a+b}-\frac{a}{(b+c)^2}-\frac{b}{(c+a)^2}\right)\delta c\tag1
\end{align}
$$
for all $\delta a,\delta b,\delta c$. That means
$$
\begin{align}
\frac1{a+b}&=\frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}\tag2\\
\frac1{b+c}&=\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\tag3\\
\frac1{c+a}&=\frac{c}{(a+b)^2}+\frac{a}{(b+c)^2}\tag4
\end{align}
$$
Subtract $(4)$ from the sum of $(2)$ and $(3)$:
$$
\frac1{b+c}+\frac1{a+b}-\frac1{c+a}=\frac{2b}{(c+a)^2}\tag5
$$
Add $\frac2{c+a}$ and divide by $2(a+b+c)$:
$$
\frac1{2(a+b+c)}\left(\frac1{b+c}+\frac1{a+b}+\frac1{c+a}\right)=\frac1{(c+a)^2}\tag6
$$
By symmetry,
$$
\frac1{(a+b)^2}=\frac1{(b+c)^2}=\frac1{(c+a)^2}\tag7
$$
from which we get $a=b=c$. Thus, we get
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac32\tag8
$$
