Nesbitt's inequality

Idea. Put every fraction in the form $\frac{x^2}{y}$.

Titu.
$$\sum\frac a{b+c}=\sum\frac{a^2}{a(b+c)}\ge\frac{(a+b+c)^2}{\sum a(b+c)}=\frac{(a+b+c)^2}{2(ab+bc+ca)}.$$

Last step. $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\ge3(ab+bc+ca)$ because $a^2+b^2+c^2\ge ab+bc+ca$ (it is $\frac12\sum(a-b)^2\ge0$). Therefore
$$\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge\frac{3(ab+bc+ca)}{2(ab+bc+ca)}=\frac32.$$

Equality iff $a=b=c$.

Remark (Titu from Cauchy-Schwarz). $\left(\sum\frac{x_i^2}{y_i}\right)\left(\sum y_i\right)\ge\left(\sum x_i\right)^2$ is Cauchy-Schwarz applied to the vectors $\left(\frac{x_i}{\sqrt{y_i}}\right)$ and $\left(\sqrt{y_i}\right)$.

Lesson. A sum of fractions $\frac{\text{something}}{\text{something}}$: multiply numerator and denominator to create squares in the numerators, then use Titu.