Prove this inequality $\sum_{cyc}\frac{a}{ab+1}\geq \frac{3}{2}$

Question

For $a>0$,$b>0$ and $c>0$ such tat $a+b+c=3$ prove that: $$\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq \frac{3}{2}$$

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By AM-GM $$\frac{a}{ab+1}=a-\frac{a^2b}{ab+1}\ge a-\frac{a^2b}{2\sqrt{ab}}=a-\frac{a\sqrt{ab}}{2}$$

$$\Rightarrow L.H.S\ge a+b+c-\frac{a\sqrt{ab}+b\sqrt{bc}+c\sqrt{ca}}{2}$$

$$\ge 3-\frac{a^2+b^2+c^2+ab+bc+ca}{4}$$

Help me to continue

Answer 1

By Vasc's inequality $(x^2+y^2+z^2)^2\geq3(x^3y+y^3z+z^3x)$ and by AM-GM we obtain: $$\sum_{cyc}\frac{a}{ab+1}=a+b+c+\sum_{cyc}\left(\frac{a}{ab+1}-a\right)=$$ $$=3-\sum_{cyc}\frac{a^2b}{ab+1}\geq3-\sum_{cyc}\frac{a^2b}{2\sqrt{ab}}=3-\frac{1}{2}\sum_{cyc}\sqrt{a^3b}\geq3-\frac{3}{2}=\frac{3}{2}.$$ A proof of the Vasc's inequality. $$(x^2+y^2+z^2)^2-3(x^3y+y^3z+z^3x)=\frac{1}{2}\sum_{cyc}(x^2-y^2-xy-xz+2yz)^2\geq0.$$