# 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.

Without loss of generality, assume that $x \geq y \geq z$. It is easy to verify that using $(x + \frac{z}{2}, y + \frac{z}{2}, 0)$ to replace $(x, y, z)$ would make the LHS smaller, i.e., $$\frac{1}{\sqrt{x^2 + y^2}} + \frac{1}{\sqrt{y^2 + z^2}} + \frac{1}{\sqrt{x^2 + z^2}} \geq \frac{1}{\sqrt{\left(x + \frac{z}{2}\right)^2 + \left(y + \frac{z}{2}\right)^2}} + \frac{1}{y + \frac{z}{2}} + \frac{1}{x + \frac{z}{2}}.$$ This is obvious as $\left( y + \frac{z}{2}\right)^2 \geq (y + z)^2, \left( x + \frac{z}{2}\right)^2 \geq (x + z)^2$.

As a consequence, we only need to prove the case where $z = 0$, i.e., for nonnegative $x$ and $y$ with $x + y = 2$,

$$\frac{1}{\sqrt{x^2 + y^2}} + \frac{1}{x} + \frac{1}{y} \geq 2 + \frac{1}{\sqrt{2}}.\tag{*}$$

Note that the LHS in (*) could be written as \begin{align} \frac{1}{\sqrt{(x+y)^2 - 2xy}} + \frac{x+y}{xy} &= \frac{1}{\sqrt{4 - 2xy}} + \frac{2}{xy}\\ &= 2 + \frac{1}{\sqrt{2}} + (1 - xy) \left[\frac{2}{xy} - \frac{1}{\sqrt{2}(2 - xy + \sqrt{2 - xy})}\right]\\ &\geq 2 + \frac{1}{\sqrt{2}} + (1 - xy)\left(2 - \frac{1}{2\sqrt{2}}\right)\\ &\geq 2 + \frac{1}{\sqrt{2}}, \end{align}

where we have used the fact that $xy \leq \left(\frac{x + y}{2}\right)^2 = 1$.

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).