<s>[INST] If $x+y+z=2$ prove that $\sum_{cyc}\frac{1}{\sqrt{x^2+y^2}}\ge2+\frac{1}{\sqrt{2}}$ Let $x,y,z$ be non-negative reals whose sum is $2$. Prove that
$\frac{1}{\sqrt{x^2+y^2}}+\frac{1}{\sqrt{y^2+z^2}}+\frac{1}{\sqrt{z^2+x^2}}\ge2+\frac{1}{\sqrt{2}}$
I have tried bounding them up (assuming that $a\le b\le c$), many inequalities (AM-GM, QM-AM, Cauchy) but nothing has worked. I know that equality is achieved when one of $x,y,z$ is $0$ and the other two are $1$ but that doesn't help.
 [/INST] The hint.
By Holder
$$\left(\sum_{cyc}\frac{1}{\sqrt{x^2+y^2}}\right)^2\sum_{cyc}(x^2+y^2)(x^2+y^2+3z^2+2(\sqrt2-1)xy)^3\geq$$
$$\geq\left(\sum_{cyc}(5x^2+2(\sqrt2-1)xy)\right)^3.$$
Thus, it remains to prove that
$$(x+y+z)^2\left(\sum_{cyc}(5x^2+2(\sqrt2-1)xy)\right)^3\geq$$
$$\geq2(9+4\sqrt2)\sum_{cyc}(x^2+y^2)(x^2+y^2+3z^2+2(\sqrt2-1)xy)^3,$$
which is true by $uvw$ (https://math.stackexchange.com/tags/uvw/info).
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