I have an inequality problem :
Let $a,b,c$ are real numbers such that $a, b, c \geq 0$ and $ab + bc + ca = 1$. Prove that $\sum \dfrac{3ab+1}{a+b} \geq 4$
Solution : $$\sum \dfrac{3ab+1}{a+b} = \sum \dfrac{4ab+bc+ca}{a+b} = \sum \dfrac{4ab}{a+b} + \sum a \geq \dfrac{4(ab+bc+ca)^{2}}{ab(a+b)+bc(b+c)+ca(c+a)} + \sum a = \dfrac{4}{(a+b+c)(ab+bc+ca)-abc} + \sum a \geq \dfrac{4}{\sum a} + \sum a \geq 4$$
Manual testing shows that equality holds on when (a, b, c) = (1; 1; 0). However if a number equals 0, WLOG it is $a$, the Cauchy-Schwarz inequality holds on when $\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}$, making $b = 0$ or $c = 0$ too. Hence, $ab + bc + ca = 0$, a contracdition. Any explanation ?
You used the following C-S: $$\sum_{cyc}ab(a+b)\sum_{cyc}\frac{ab}{a+b}\geq(ab+ac+bc)^2.$$ The equality occurs, when $$\left(\sqrt{ab(a+b)},\sqrt{bc(b+c)},\sqrt{ca(c+a)}\right)||\left(\sqrt{\frac{ab}{a+b}},\sqrt{\frac{bc}{b+c}},\sqrt{\frac{ca}{c+a}}\right),$$ which for $(a,b,c)\rightarrow(1,1,0)$ gives $$(\sqrt2,0,0)||\left(\frac{1}{\sqrt2},0,0\right),$$ which is true.