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.

We need to prove that
$$\sum_{cyc}\sqrt{\frac{x^2+y^2}{2}}\geq\sqrt2$$ or
$$\sum_{cyc}\sqrt{x^2+y^2}\geq2,$$
which is true because
$$\sum_{cyc}\sqrt{x^2+y^2}=\sqrt{\sum_{cyc}(x^2+y^2+2\sqrt{(x^2+y^2)(x^2+z^2)})}\geq$$
$$\geq\sqrt{\sum_{cyc}(x^2+y^2+2x^2)}=2$$
