<s>[INST] Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality 
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$ 
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc.
Using that I get this
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$ 
From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering that $a+b\geq 2\sqrt{ab}$ 
But by using this, I get the following...
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}} \leq 3 \times \sqrt[3]{\dfrac{abc}{8abc}} = \dfrac{3}{2}$ 
Everything seems so perfect because I get the value $\dfrac{3}{2}$ as required, but this method isn't valid due to the change in direction! What is going on?
Is there a way of proving this inequality otherwise then?
 [/INST] If we denote $s:=a+b+c$, we have to minimize
$$\frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}$$
for $a,b,c>0$.
Using new variables $x:=a/s$, $y:=b/s$, $z:=c/s$ we can see, that this is equivalent to minimizing
$$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}$$
for $x+y+z=1$.
If we notice that the function
$$f(x)=\frac{x}{1-x}=-1+\frac1{1-x}=-1-\frac1{x-1}$$
is convex on the interval $(0,1)$, we have
$$\frac{f(x)+f(y)+f(z)}3 \ge f\left(\frac{x+y+z}3\right),$$
i.e.
$$f(x)+f(y)+f(z) \ge 3f(1/3).$$
By computing $f(1/3)=1/2$ we see, that the last inequality is precisely
$$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z} \ge \frac32.$$
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