Given $~ \mathrm{a,b,c >0 ~ , ~ a+b+c=1 } ~ $ then prove $~ \mathrm{\sum\limits_{cyc} \sqrt{a+b^2 } \geqslant 2 } $

Given $~ \mathrm{a,b,c >0 ~ , ~ a+b+c=1 } ~ $ then prove $~ \mathrm{\sum\limits_{cyc} \sqrt{a+b^2 } \geqslant 2 } $

Here's what I've tried : $\sum\limits_{cyc} \sqrt{a+b^2 } = \sum\limits_{cyc} \sqrt{\sqrt a^2 +b^2 }=\sum\limits_{cyc}\sqrt{\frac{1}{2}(\sqrt a^2 +b^2 )(1^2 +1^2 ) } \ge \\ \ge \sum\limits_{cyc} \sqrt{\frac{1}{2} (\sqrt{a} +b )^2 }=\frac{1}{\sqrt 2} (1+ \sqrt a +\sqrt b + \sqrt c ). $

Any ideas how to proceed and am I on the right track ?

COMMENT.- A method to solve these inequalities graphically. Leaving fixed $a$, we have to study the relative position of the line $L: x + y = 1-a$ with respect to the curve $$\Gamma: \sqrt{a+x^2}+\sqrt{x+y^2}+\sqrt{y+a^2}=2$$ where $0\lt a, x, y \lt1$. which is decreasing in the first quadrant because $\dfrac{dy}{ dx}\lt 0$.

$L$ is tangent to $\Gamma$ at the point $(\frac13,\frac13)$ and it is the only point where it touches the curve ($L$ cannot cut the curve in the first quadrant because if it does then the inequality is false in a certain interval $(x_1, x_2)$ , in other words the line $L$ should be always in top of $\Gamma$ , at most it can be tangent to $\Gamma$). The calculations can be hard in some cases. Anyway this is what happen for all $a$ and it is easily verified graphically.