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Taken from Titu Andreescu and Bogdan Enescu's Mathematical Olympiad Treasures on page 9 Problem 1.19, to prove,
$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geqslant \frac{3}{2}$.
It is easy to see why $LHS$ may yield
$\geqslant\frac{(a+b+c)^2}{2(ab+bc+ca)}$ from Cauchy.
Yet how do one bring in the '$3$' part from the $RHS$ knowing that $a^2+b^2+c^2 \geqslant ab+bc+ca $?
Any help would be appreciated.
Because $$\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2}$$ it's $$(a+b+c)^2\geq3(ab+ac+bc)$$ or $$a^2+b^2+c^2+2ab+2ac+2bc\geq3(ab+ac+bc)$$ or $$a^2+b^2+c^2\geq ab+ac+bc.$$
Write $x:=b+c,\,y:=c+a,\,z:=a+b$ so $\frac{a}{b+c}=\frac{y+z-x}{2x}$. The desired result is equivalent to $\sum_\text{cyclic}\frac{y+z}{x}\ge6$, which is trivial because Cauchy-Schwarz implies $q+\frac1q\ge2$ (hint: use $\binom1q,\,\binom{q}{1}$) for $q\in\{\frac{x}{y},\,\frac{y}{z},\,\frac{z}{x}\}$.
