Key idea. Apply C4.3 twice, then use a sum of squared differences.

1. Let \(s=a+b+c\) and \(q=ab+bc+ca>0\). Rewrite the target left side as

   \[L=\frac{a^2}{a(b+c)}+\frac{b^2}{b(c+a)}+\frac{c^2}{c(a+b)}.\]

2. Put \(D=a(b+c)+b(c+a)>0\). Applying C4.3 to the first two fractions gives

   \[L\ge\frac{(a+b)^2}{D}+\frac{c^2}{c(a+b)}.\]

3. Apply C4.3 once more, now to these two fractions. The sum of the denominators is \(D+c(a+b)=2q\), so

   \[L\ge\frac{(a+b+c)^2}{D+c(a+b)}=\frac{s^2}{2q}.\]

4. Finally,

   \[\begin{aligned}s^2-3q&=a^2+b^2+c^2-ab-bc-ca\\&=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\ge0.\end{aligned}\]

   Thus \(s^2\ge3q\), and division by \(2q>0\) gives \(L\ge3/2\).

5. If \(a=b=c\), each original fraction equals \(1/2\), so equality holds. Conversely, equality in the final bound forces all three squared differences to vanish, so \(a=b=c\) is necessary.

Remember. A proof can combine a fraction inequality with a separate bound on a symmetric expression.