<s>[INST] For positive $a$, $b$, $c$ with $a+b+c=1$, show that $(ab+bc+ca) \sum_{cyc}\frac{a}{b^2+b} \geq \frac34$ 
If $a,b,c > 0$ and $a+b+c = 1$, then prove that
  $$\left(\frac{a}{b^2+b}+\frac{b}{c^2+c}+\frac{c}{a^2+a}\right)(ab+bc+ca)\geq\frac{3}{4}$$

It's been more than 35 years since I last touched algebra!! 
 [/INST] Hint: Using generalised Holder’s inequality
$$\sum_{cyc}ab \cdot \sum_{cyc} \frac{a}{b^2+b}\cdot \sum_{cyc} a(b+1) \geqslant \left(\sum_{cyc}a\right)^3=1$$
Now it is enough to show $\sum_{cyc} a(b+1) \leqslant \frac43$ which is easy.
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