Show that $x(t)<0$ for all $t>0$

By the mean value theorem, we have $x(t)-x(0)=x'(r)(t-0)$ for some $0<r<t$. Substituting values, we get $x(t)+1=tx'(r)$. Applying the hypothesis, we have now that $x(t)+1\le tx(r)^2$. But $x'$ is positive everywhere, so $x$ is increasing, which means that $x(t)+1\le tx(r)^2\le tx(t)^2$ and so $tx(t)^2-x(t)-1\ge 0.$ Noting that this inequality implies that there is no $t\ge 0$ for which $x(t)=0$, and since $x(0)=-1$, we may invoke the intermediate value theorem to conclude that $x(t)<0$ on $[0,\infty).$