I'm having difficulties in proving that the projection of $$(s,y)\in R \times R^{n}$$ onto the second-order cone $$Q^{n+1} = \{(t,x) \in R \times R^n : \|x\|_2 \leq t \}$$ is $$ \frac{s+\|y\|_2}{2\|y\|_2} (\|y\|_2,y)$$ when $\|y\|_2 > s $ and $ \|y\|_2 > -s $.
I tried to first show that to minimize the distance, $x$ must be parallel to $y$, then I construct $$\alpha (\|y\|_2 + \epsilon,y)$$ and minimize the distance over $\alpha$ and $\epsilon$.
I indeed come up with two quadratic functions individually attains its minimum when $$\epsilon = 0$$ and $$\alpha = \frac{s+\|y\|_2}{2\|y\|_2}$$ but I still have $$-\frac{s^2(\|y\|_2+\epsilon)}{2\|y\|_2} - \frac{{\|y\|_2}^3}{2\|y\|_2 + \epsilon}$$ which attains its maximum when $$\epsilon = 0$$ As a result, I can't say the distance attains its minimum accordingly. Is there any other method or elementary method to prove the optimal solution is $$ \frac{s+\|y\|_2}{2\|y\|_2} (\|y\|_2,y)$$ when $\|y\|_2 > s $ and $ \|y\|_2 > - s $ ?
Let $(t^*,x^*)$ be the projection. The optimality condition implies that for any $(t,x) \in Q^{n+1}$ we must have $$\langle(t-t^*, x-x^*), (s-t^*, y-x^*)\rangle \le 0.$$ Applying this inequality to $(t,x)=(0,0)$ and $(t,x) = 2(t^*,x^*)$ shows that in particular we have $$\langle(t^*, x^*), (s-t^*, y-x^*)\rangle = 0\tag{1}$$ and consequently, the first inequality can be rewritten as $$\langle(t,x), (s - t^*, y - x^*)\rangle \le 0.\tag{2}$$ These last two lines actually characterize $(t^*,x^*)$. That is, $(t^*,x^*)$ is the unique element of $Q^{n+1}$ satisfying both (1) and (2).
Note that $(s-t^*, y -x^*) = \frac{\|y\|_2 - s}{2} (-1, y/\|y\|_2)$. From here (1) is easily verified. To verify $2$, note that the left-hand side is $$\frac{\|y\|_2 - s}{2} (-t + x^\top y / \|y\|_2),$$ and the first term is positive by the assumption $\|y\|_2 > s$, and the second term is nonpositive by Cauchy-Schwarz and the definition of $Q^{n+1}$.