Show that $Y_t=(t+1)X_{\frac{t}{t+1}}$ is Brownian motion if $X_t$ is Brownian motion.

Question

So I have to check if $Y_t$ is Wiener process first:

$$\operatorname{E}[Y_t]=0$$

$$\operatorname{Cov}(Y_s,Y_t)=\min\{s,t\}$$

First condition is easy, but with the second i came across the problem: $$ \operatorname{Cov}(Y_s,Y_t)=\operatorname{E}[Y_sY_t] = \operatorname{E}\left[ (s+1) X_{\frac{s}{s+1}}(t+1) X_{\frac{t}{t+1}} \right] = (t+1)(s+1) \operatorname{E} \left[X_{\frac{t}{t+1}}X_{\frac{s}{s+1}}\right]$$

If $X_t$ is Brownian motion then $E[X_sX_t]=\min\{s,t\}$? But if i suppose that, for example, $s

What am I doing wrong?