<s>[INST] Inequality. $\frac{ab+c}{a+b}+\frac{ac+b}{a+c}+\frac{bc+a}{b+c} \geq 2.$ Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that (using rearrangements inequalities, you can also view this exercise here, exercise number 3.1.8 )
$$\frac{ab+c}{a+b}+\frac{ac+b}{a+c}+\frac{bc+a}{b+c} \geq 2.$$
thanks. 
 [/INST] The inequality follows $$\frac{(b+c)(c+a)}{a+b}+\frac{(a+c)(a+b)}{b+c}+\frac{(a+b)(b+c)}{a+c}\geq 2$$
Let $a+b=z,b+c=x,c+a=y$ then $x+y+z=2$. Therefore $$x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}\geq xyz(x+y+z)$$
This is true by AM-GM.
P/s: Sorry for my bad English.
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