<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] We can also use rearrangement inequality to prove it.  
$\displaystyle\frac a{b+c}+\frac b{a+c}+\frac c{a+b}$
$\displaystyle=\frac12\left(\left(\frac a{b+c}+\frac b{a+c}+\frac c{a+b}\right)+\left(\frac a{b+c}+\frac b{a+c}+\frac c{a+b}\right)\right)$
$\displaystyle\geq\frac12\left(\left(\frac a{a+c}+\frac b{a+b}+\frac c{b+c}\right)+\left(\frac a{a+b}+\frac b{b+c}+\frac c{a+c}\right)\right)$
$=\dfrac32$
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