Nesbitt's inequality via Cauchy-Schwarz

Engel form.
$$\sum_{\text{cyc}}\frac{a}{b+c}=\sum_{\text{cyc}}\frac{a^2}{a(b+c)}\ge\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}.$$

Denominator.
$$a(b+c)+b(c+a)+c(a+b)=2(ab+bc+ca).$$

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

Conclude. Using $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\ge3(ab+bc+ca)$ (since $a^2+b^2+c^2\ge ab+bc+ca$), we get
$$\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge\frac{3(ab+bc+ca)}{2(ab+bc+ca)}=\frac32.$$

Equality. $a=b=c$.