Let $u$ solve $u_t - \Delta u + u^2 = 0$ on $[0,T] \times\mathbb{R}^d$. Prove that $u(T,x) \leq 1/T$ for all $x$ (regardless of initial conditions).
My attempt: We see that $u_t - \Delta u = -u^2 \leq 0$ on $\overline{U_T}$. Hence, by the maximum principle, we see that $\max_{\partial pU_T}u = \max_{U_T} u $. How can I use this fact to prove the bound on $u(T,x)$? It seems as if I have too little information about $u$ to do so. I was thinking of constructing some auxiliary function and applying the comparison principle, but it seems that not even this will work.
Observe that $v(t)=1/t$ satisfies the equation $v_t-\Delta v+v^2=0$. For all $x\in[0,1]$, $\lim_{t\to0^+}v(t)>u(x,0)$. By the comparison principle, $u(x,t)\le v(t)$.