Is the inequality $\sqrt{(2^r + 1)(t + 1)} \leq \min(2^r, t)$ false in general?

Question

My question here is pretty basic:

Is the inequality $\sqrt{(2^r + 1)(t + 1)} \leq \min(2^r, t)$ false in general?

I tried asking WolframAlpha for its solutions, it says that no solutions exist.

I also tried asking WolframAlpha for the solutions to the closely related inequality $$\sqrt{(2^r + 1)(t + 1)} \leq \max(2^r, t),$$ it gave the following solutions: $$t \geq \frac{\sqrt{2^{2r} + 3 \times {2^{r+1}} + 5} + 2^r + 1}{2}$$ and $$-1 \leq t \leq \frac{-2^r + 2^{2r} - 1}{2^r + 1}.$$

Note that the first solution is for $t > 2^r$ (and therefore, $\max(2^r, t) = t$), and that the second solution is for $t < 2^r$ (and therefore, $\max(2^r, t) = 2^r$).

Finally, here is the inequality plot for $$\sqrt{(2^r + 1)(t + 1)} \leq \max(2^r, t):$$

Answer 1

$$\sqrt{(2^r+1)(t+1)} \le \min(2^r, t)$$

The LHS is nonnegative, hence $t$ is nonnegative.

\begin{align} \sqrt{(2^r+1)(t+1)} &\ge \sqrt{(\min(2^r, t)+1)(\min(2^r,t)+1) } \\&= \sqrt{(\min(2^r, t)+1)^2} \\&= \min(2^r,t)+1 \\&>\min(2^r, t) \ \end{align}