What technique can be used to prove this inequality? For $a,b\in(0,1)$, prove
$$\sqrt{a^2+b^2}+\sqrt{\frac{1}{4}+\left(1-a\right)^2}+\sqrt{\frac{4}{9}+\left(1-b\right)^2}\ge \frac{\sqrt{181}}{6}$$
where the equality holds for $a=\frac49$ and $b=\frac25$.
Attempted for a while, it seems no standard inequality tricks can be utilized at all in stead of solving the complicated derivative equation group.
$$\frac{d}{da}\left(\sqrt{a^2+b^2}+\sqrt{\frac{1}{4}+\left(1-a\right)^2}+\sqrt{\frac{4}{9}+\left(1-b\right)^2}\right)=\frac{a}{\sqrt{a^2 + b^2}}-\frac{2(1 - a)}{\sqrt{4 a^2 - 8 a + 5}}$$
and
$$\frac{d}{db}\left(\sqrt{a^2+b^2}+\sqrt{\frac{1}{4}+\left(1-a\right)^2}+\sqrt{\frac{4}{9}+\left(1-b\right)^2}\right)=\frac{b}{\sqrt{a^2 + b^2}}-\frac{1 - b}{\sqrt{(1 - b)^2 + \frac49}}.$$

If you're looking for 'standard inequalities', then check Minkowski's inequality,
$$\sqrt{\color{red}{a^2}+\color{blue}{b^2}} + \sqrt{\color{red}{(1-a)^2}+\color{blue}{\tfrac14}} + \sqrt{\color{red}{\tfrac49}+\color{blue}{(1-b)^2}} \geqslant \sqrt{\color{red}{(a+1-a+\tfrac23)^2} + \color{blue}{\left(b+\tfrac12+1-b \right)^2}} \\= \sqrt{\color{red}{\frac{25}9}+\color{blue}{\frac94}}$$

P.S. Maybe simpler to rephrase this in terms of the triangle inequality, if you can see an appropriate choice of points.
