Prove that $\sum_{cyc}\sqrt{a^2+\frac{1}{\left(b+1\right)^2}}\ge \frac{\sqrt{181}}{5}$

Question

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

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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

Answer 1

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!