2. Note that $ab+bc+ca\ge 3\sqrt[3]{(abc)^2}$, so $abc\le 1$. Then, by Cauchy: \begin{align*} \sum_{cyc} \dfrac{ab}{c+1} &=\sum_{cyc}\dfrac{(ab)^2}{abc+ab}\\ &\ge \dfrac{(ab+bc+ac)^2}{3abc+ab+bc+ca}\\ &=\dfrac{9}{3abc+3}\\ &\ge \dfrac{3}{2} \end{align*}

rogue wrote: 1) Let $a,b,c\ge0$ and $ab+bc+ca=2.$ Prove that \[\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}+2(a+b+c)\ge6.\]2) Let $a,b,c\ge0$ and $ab+bc+ca=3.$ Prove that \[\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}\ge\frac{3}{2}.\] We have 1) \[LHS-RHS=\sum{\frac{(a-b)^2}{(a+1)(b+1)}}\ge{0}\] 2) \[LHS-RHS=\frac{\sum{c(2c+3)(a-b)^2}}{6(a+1)(b+1)(c+1)}\ge{0}\]

rogue wrote: 2) Let $a,b,c\ge0$ and $ab+bc+ca=3.$ Prove that \[\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}\ge\frac{3}{2}.\] $\begin{array}{l} \frac{{ab}}{{c + 1}} + \frac{{bc}}{{a + 1}} + \frac{{ca}}{{b + 1}} \ge \frac{3}{2} \Leftrightarrow 2\left[ {\sum {{a^2}{b^2}} + \sum {ab\left( {a + b} \right)} + \sum {ab} } \right] \ge 3\left( {\sum a + \sum {ab} + abc + 1} \right)\\ \Leftrightarrow 2\sum {{a^2}{b^2}} + 2\sum {ab\left( {a + b} \right)} \ge 3\sum a + 3abc + 6\\ \Leftrightarrow 2\sum {{a^2}{b^2}} + 2\left( {\sum a .\sum {ab} - 3abc} \right) \ge 3\sum a + 3abc + 6\\ \Leftrightarrow 2\sum {{a^2}{b^2}} + 6\sum a \ge 3\sum a + 9abc + 6\\ \Leftrightarrow 2\sum {{a^2}{b^2}} + 3\sum a \ge 9abc + 6 \end{array}$ It's true by $2\sum {{a^2}{b^2}} \ge 2\frac{{{{\left( {\sum {ab} } \right)}^2}}}{3} = 6{\rm{ and }}3\sum a = \sum a .\sum {ab} \ge 9abc.{\rm{ Done!}}$.

1)we know from condition $\frac{2\sqrt{2}}{3\sqrt{3}}\geq abc.$ Then ,we know $C-S:$ $$\sum_{cyc} \frac{ab}{c+1} +2(a+b+c)\geq \frac{(ab+bc+ca)^2}{3abc+ab+bc+ca}+2(a+b+c)=\frac{4}{2+3abc}+2(a+b+c)\geq 6.$$Then,we know $a+b+c=\sqrt{a^2+b^2+c^2+4}\geq \sqrt{ab+bc+ca+4}=\sqrt{6}.$ Then the equivalent inequality is : $\frac{2}{2+3abc}+\sqrt{6}\geq \frac{2}{2+\frac{2\sqrt{2}}{\sqrt{3}}}+\sqrt{6}=3-\sqrt{6}+\sqrt{6}=3.$

for the second inequality I have a simple solition: the function $f:(0,\infty )\to (0,\infty )$, $f(x)=\frac{1}{x+1}$ it is a convex one. Pick: $p_1=\frac{ab}{3}$, $p_2=\frac{bc}{3}$,$p_1=\frac{ca}{3}$ and $x_1=c$, $x_2=a$,$x_1=b$. Apply $\sum_{i=1}^3 p_i f(x_i)\geq f(p_i x_i)$ and we are done.

kukyrami wrote: for the second inequality I have a simple solution: the function $f:(0,\infty )\to (0,\infty )$, $f(x)=\frac{1}{x+1}$ it is a convex one. Pick: $p_1=\frac{ab}{3}$, $p_2=\frac{bc}{3}$,$p_1=\frac{ca}{3}$ and $x_1=c$, $x_2=a$,$x_1=b$. Apply $\sum_{i=1}^3 p_i f(x_i)\geq f(p_i x_i)$ and we are done.

1) $\frac{ab}{c+1}+(a+b)=\frac{ab+ac+bc+a+b}{c+1}=\frac{a+b+2}{c+1}=\frac{a+1}{c+1}+\frac{b+1}{c+1}$ So $\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}+2(a+b+c) = \frac{a+1}{b+1}+\frac{b+1}{a+1}+\frac{c+1}{b+1}+\frac{b+1}{c+1}+\frac{a+1}{c+1}+\frac{c+1}{a+1}\geq 6$

RagvaloD wrote: 1) $\frac{ab}{c+1}+(a+b)=\frac{ab+ac+bc+a+b}{c+1}=\frac{a+b+2}{c+1}=\frac{a+1}{c+1}+\frac{b+1}{c+1}$ So $\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}+2(a+b+c) = \frac{a+1}{b+1}+\frac{b+1}{a+1}+\frac{c+1}{b+1}+\frac{b+1}{c+1}+\frac{a+1}{c+1}+\frac{c+1}{a+1}\geq 6$ Nice.