Without loss of generality, we assume that $x\ge y\ge z$.
Lemma 1:
$$\dfrac{1}{\sqrt{y^2+1}}+\dfrac{1}{\sqrt{z^2+1}}\ge 1+\dfrac{1}{\sqrt{(y+z)^2+1}}$$
Proof: this inequality is equivalent to
$$\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}+\dfrac{2}{\sqrt{(y^2+1)(z^2+1)}}\ge 1+\dfrac{1}{(y+z)^2+1}+\dfrac{2}{\sqrt{(y+z)^2+1}}.$$
Notice that
$$(y+z)^2+1-(y^2+1)(z^2+1)=yz(2-yz)\ge 0,$$
hence
$$\dfrac{2}{\sqrt{(y^2+1)(z^2+1)}}\ge\dfrac{2}{\sqrt{(y+z)^2+1}}.$$
Therefore, it suffices to prove that
$$\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}\ge1+\dfrac{1}{(y+z)^2+1}.$$
And this is equivalent to
$$\dfrac{yz[2-2yz-yz(y+z)^2]}{(y^2+1)(z^2+1)[(y+z)^2+1]}\ge 0.$$
Above is true because
$$2-2yz-yz(y+z)^2=2x(y+z)-yz(y+z)^2=(y+z)[2x-yz(y+z)]\ge (y+z)[2x-x^2(y+z)]=x(y+z)(2-xy-xz)\ge 0.$$
Lemma 2:
$$\dfrac{1}{x+y}+\dfrac{1}{\sqrt{x+z}}\ge\sqrt{y+z}+\sqrt{\dfrac{y+z}{(y+z)^2+1}}.$$
Proof:
\begin{align}\dfrac{1}{x+y}+\dfrac{1}{\sqrt{x+z}}&=\sqrt{y+z}\left(\dfrac{1}{\sqrt{(y+z)(x+y)}}+\dfrac{1}{\sqrt{(z+x)(z+y)}}\right)\\
&=\sqrt{y+z}\left(\dfrac{1}{\sqrt{1+y^2}}+\dfrac{1}{\sqrt{1+z^2}}\right)
\end{align}
and apply Lemma 1, then done!
Thus the original inequality is equivalent to
$$\dfrac{1}{\sqrt{y+z}}+\sqrt{y+z}+\sqrt{\dfrac{y+z}{(y+z)^2+1}}\ge 2+\dfrac{1}{\sqrt{2}}.$$
Letting $$t=\dfrac{1}{\sqrt{y+z}}+\sqrt{y+z}\ge 2,$$ the inequality becomes
$$t+\dfrac{1}{\sqrt{t^2-2}}\ge 2+\dfrac{1}{\sqrt{2}}.$$
This can be shown by the following:
\begin{align}
&t+\dfrac{1}{\sqrt{t^2-2}}-2-\dfrac{1}{\sqrt{2}}
\\
=&t+\dfrac{2\sqrt{2}}{2\sqrt{2}\cdot\sqrt{t^2-2}}-2-\dfrac{1}{\sqrt{2}}
\\
\ge& t+\dfrac{2\sqrt{2}}{2+t^2-2}-2-\dfrac{1}{\sqrt{2}}
\\
=&t+\dfrac{2\sqrt{2}}{t^2}-2-\dfrac{1}{\sqrt{2}}
\\
=&\dfrac{(t-2)(\sqrt{2}t^2-t-2)}{\sqrt{2}t^2}\ge 0.
\end{align}
Best Answer
Well,I think you can use the famouse Iran 96 inequality as a result to solve this problem. the Iran 96 inequality
Let $x,y,z\geq 0$ we have
$$ \frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(x+z)^{2}}\geq \frac{9}{4(xy+yz+zx)} $$ square both side,we can rewrite the inequality into $$ \sum_{cyc}{\frac{1}{(x+y)^{2}}}+2\sum_{cyc}{\frac{1}{(x+y)(x+z)}}\geq \frac{25}{4}$$ Now,Using this inequality as a known result,it's suffice to prove $$ \sum_{cyc}{\frac{1}{(x+y)(x+z)}}\geq 2 $$ after simple homogenous,it's $$ (xy+yz+xz)(x+y+z)\geq (x+y)(y+z)(z+x) $$ Or $$ xyz\geq 0 $$ Which is obviously true,Equality occurs if and only if $x=y=1, z=0 $ or it's permutation. The proof is complete