<s>[INST] Prove by this inequality to with $x^2+y^2+z^2=1$ Let $x,y,z>0,x^2+y^2+z^2=1$ show that
$$\sum_{cyc}\sqrt{\dfrac{x^2+y}{2}}\ge\sqrt{2}$$
Here's what I have done. The expression $x^2+y\ge x^2+y^2$. it is  is equivalent to
$$\sum_{cyc}\sqrt{1-x^2}\ge 2$$ shows that the function is neither concave or convex. So I don't think Jensen useful here.
 [/INST] You have
$$
2 = \sum_{cyc} (x^2 + y^2) \le \sum_{cyc}  \sqrt{x^2 + y^2}
\le \sum_{cyc}  \sqrt{x^2 + y}
$$
because $a \le \sqrt a$ for $0 \le a \le 1$.
The inequality is sharp, equality holds exactly if $(x, y, z)$ is
any permutation of $(1, 0, 0)$.
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