Since the inequality is invariant under the scaling $(x,y)\mapsto (tx,ty)$, we can assume $\|x\|=1$. Set $r=\|y\|$ and $\alpha = \langle x,y \rangle / r \in [1,1]$. We now have the inequality $(1+r)\alpha \le \sqrt{1 + r^2 + 2 r \alpha}, \quad r \ge 0,\, \alpha \in [-1,1]$ which should be a simple calculus exercise to verify.
It is trivial when $\alpha < 0$, so let us assume $\alpha \ge 0$. In this case we can square both sides to get the equivalent inequality $(1+r)^2 \alpha^2 \le 1+r^2+2r\alpha.$ So if we set $F(r,\alpha) = 1+r^2+2r\alpha-(1+r)^2\alpha^2, \quad r \ge 0, \,\alpha \in [0,1]$ we can see $F \ge 0$ for $r = 0$, $r \to +\infty$, $\alpha = 0$ and $\alpha = 1$. It remains to look for critical points of $F$ inside the region $r > 0$, $\alpha \in (0,1)$. However, differentiating with respect to $r$, we have $F_r(r,\alpha) = 2r + 2\alpha - 2\alpha^2(1+r) = 2r(1-\alpha^2) + 2\alpha(1-\alpha)$ which is strictly positive on $r > 0$, $\alpha \in (0,1)$. So there are no critical points, and we have $F \ge 0$ on the entire region.