Not so easy

By Titu's Lemma,

$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} = \dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge \dfrac{(a+b+c)^2}{2(ab+bc+ca)}.$

Next, we prove $\begin{array}{c}&\dfrac{(a+b+c)^2}{2(ab+bc+ca)}\ge \dfrac{3}{2} &\text{ or } &(a+b+c)^2\ge 3(ab+bc+ca), \end{array}$ which is true after full expansion and the use of the inequality $a^2+b^2+c^2\ge ab+bc+ca$ .

$Let X = \dfrac a{b+c} + \dfrac b{a+c} + \dfrac c{a+b}$

$X + 3 = (a + b + c)( \dfrac 1{b+c} + \dfrac 1{a+c} + \dfrac 1{a+b} )$

Note that :

$\dfrac {(a + b) + (b + c) + (c + a)}{3} \ge \dfrac {3}{\dfrac {1}{a + b} + \dfrac{1}{b + c} + \dfrac{1}{c + a}} [AM \ge HM]$

$\dfrac {1}{a + b} + \dfrac{1}{b + c} + \dfrac{1}{c + a} \ge \dfrac{9}{2(a + b + c)}$

$X + 3 \ge (a + b + c)\dfrac{9}{2(a + b + c)}$

$X \ge \dfrac{9}{2} - 3$

$\dfrac a{b+c} + \dfrac b{a+c} + \dfrac c{a+b} \ge \dfrac{3}{2}$

Hence 3 + 2 = 5