For positive $a$, $b$, $c$ with $a+b+c=1$, show that $(ab+bc+ca) \sum_{cyc}\frac{a}{b^2+b} \geq \frac34$

Solution 1:

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.

For positive $a$, $b$, $c$ with $a+b+c=1$, show that $(ab+bc+ca) \sum_{cyc}\frac{a}{b^2+b} \geq \frac34$

Solution 1:

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