Prove that $\sum_{cyc}\sqrt{a^2+\frac{1}{\left(b+1\right)^2}}\ge \frac{\sqrt{181}}{5}$ For $a,b,c>0$ satisfy $ab+bc+ca\ge \frac{4}{3}$. Prove that $$\sqrt{a^2+\frac{1}{\left(b+1\right)^2}}+\sqrt{b^2+\frac{1}{\left(c+1\right)^2}}+\sqrt{c^2+\frac{1}{\left(a+1\right)^2}}\ge \frac{\sqrt{181}}{5}$$

My try: By Minkowski: 
$LHS\ge \sqrt{\left(a+b+c\right)^2+\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)^2}$
We have: $(a+b+c)^2\ge 3(ab+bc+ca)=3.\frac{4}{3}=4$
And by C-S: $\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)^2\ge \left(\frac{9}{a+b+c+3}\right)^2$
I can't continue, help me
 A: Let $a=\frac{2x}{3}$, $b=\frac{2y}{3}$ and $c=\frac{2z}{3}$.
Hence, the condition gives $xy+xz+yz\geq3$ and we need to prove that
$$\sum_{cyc}\sqrt{\frac{4x^2}{9}+\frac{9}{(2y+3)^2}}\geq\frac{\sqrt{181}}{5}.$$
Now, by C-S
$$\sum_{cyc}\sqrt{\frac{4x^2}{9}+\frac{9}{(2y+3)^2}}=\frac{15}{\sqrt{181}}\sum_{cyc}\sqrt{\left(\frac{4}{9}+\frac{9}{25}\right)\left(\frac{4x^2}{9}+\frac{9}{(2y+3)^2}\right)}\geq$$
$$\geq\frac{15}{\sqrt{181}}\sum_{cyc}\left(\frac{4x}{9}+\frac{9}{5(2y+3)}\right).$$
Thus, it remains to prove that
$$\frac{15}{\sqrt{181}}\sum_{cyc}\left(\frac{4x}{9}+\frac{9}{5(2y+3)}\right)\geq\frac{\sqrt{181}}{5}$$ or
$$20(x+y+z)+81\left(\frac{1}{2x+3}+\frac{1}{2y+3}+\frac{1}{2z+3}\right)\geq\frac{543}{5}.$$
Now, let $x+y+z=3u$ and $xy+xz+yz=3v^2$, where $v>0$. 
Hence, $u\geq v\geq1$ and by C-S we obtain:
$$20(x+y+z)+81\left(\frac{1}{2x+3}+\frac{1}{2y+3}+\frac{1}{2z+3}\right)-\frac{543}{5}\geq$$
$$\geq20(x+y+z)+81\cdot\frac{(1+1+1)^2}{\sum\limits_{cyc}(2x+3)}-\frac{543}{5}=60u+\frac{729}{6u+9}-\frac{543}{5}=$$
$$=3\left(20u+\frac{81}{2u+3}-\frac{181}{5}\right)=\frac{6(u-1)(100u+69)}{5(2u+3)}\geq0.$$
Done!
