<s>[INST] Given $~ {a,b,c \ge 0 ~ , ~ a+b+c=1 } ~ $ then prove $~ {\sum\limits_{cyc} \sqrt{a+b^2 } \geqslant 2 } $ 
Given  $~ a,b,c \ge 0  ~ , ~ a+b+c=1  ~ $, prove that
$$\sum\limits_{\mathrm{cyc}} \sqrt{a+b^2 } := \sqrt{a + b^2} + \sqrt{b + c^2} + \sqrt{c + a^2}\ge 2.$$

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 ?
 [/INST] WLOG, assume that $a = \max(a, b, c)$.
It suffices to prove that
\begin{align*}
 \sqrt{a + b^2} + \sqrt{a + c^2} &\ge 2a + b + c, \tag{1}\\[6pt]
 \sqrt{b + c^2} + \sqrt{c + a^2} &\ge \sqrt{a + c^2} + b + c. \tag{2}
\end{align*}
(Add them up to get the desired result.)
First, using Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
 &(\sqrt{a + b^2} + \sqrt{a + c^2})^2 - (2a + b + c)^2\\
 =\,& a + b^2 + a + c^2 + 2\sqrt{(a + b^2)(a + c^2)} - (2a + b + c)^2\\
 \ge\,& a + b^2 + a + c^2 + 2(a + bc) - (2a + b + c)^2\\
 =\,& 4a(1 - a - b - c)\\
 =\,& 0.
\end{align*}
Thus, (1) is true.
Second, we have
\begin{align*}
 &\sqrt{b + c^2} + \sqrt{c + a^2} - \sqrt{a + c^2} - b - c\\
 =\,& [\sqrt{b + c^2} - (b + c)] - (\sqrt{a + c^2} - \sqrt{c + a^2})\\
 =\,& \frac{b(1 - b - 2c)}{\sqrt{b + c^2} + b + c} - \frac{(a - c)(1 - a - c)}{\sqrt{a + c^2} + \sqrt{c + a^2}}\\
 =\,& \frac{b(a - c)}{\sqrt{b + c^2} + b + c} - \frac{(a - c)b}{\sqrt{a + c^2} + \sqrt{c + a^2}}\\
 \ge\,& 0
\end{align*}
where we have used $\sqrt{a+c^2} \ge \sqrt{b+c^2}$ and
$(c+a^2) - (b+c)^2 = (c+a^2) - (1-a)^2 = c + 2a - 1 = a - b \ge 0$ (so $\sqrt{c+a^2} \ge b + c$).
Thus, (2) is true.
We are done.
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