<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] Observe $a + b = 1 - c$, $a + c = 1 - b$, and $b + c = 1 - a$, so the desired inequality is
$$\frac{ab+c}{1 - c}+\frac{ac+b}{1 - b}+\frac{bc+a}{1 - a} \geq 2$$
Similarly, we substitute $c = 1 - a - b$, $b = 1 - a - c$, and $a = 1 - b - c$ in the numerator, and the desired inequality becomes
$$\frac{ab + 1 - a - b }{1 - c}+\frac{ac+1 - a - c}{1 - b}+\frac{bc+ 1 - b - c}{1 - a} \geq 2$$
This can be rewritten as
$$\frac{(1 - a)(1-b)}{1 - c}+\frac{(1 - a)(1 - c)}{1 - b}+\frac{(1 - b)(1 - c)}{1 - a} \geq 2$$
It's natural to let $A = 1 - a$, $B = 1 - b$, and $C = 1 - c$ here. So we want to show under the condition that $A + B + C = 2$ that we have the following.
$$\frac{AB}{C}+\frac{AC}{B}+\frac{BC}{A} \geq 2 {\hspace 1 in}(*)$$
Without loss of generality, we may assume $A \leq B \leq C$. Then the rearrangement inequality says the left-hand side of $(*)$ is at least as large of what you get by any permutation of the denominator. So you have
$$\frac{AB}{C}+\frac{AC}{B}+\frac{BC}{A} \geq \frac{AB}{A}+\frac{AC}{C}+\frac{BC}{B}$$
$$ = B + A + C$$
$$ = 2$$
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