Can we say $\sqrt {ab} \ge \min \{ a,b\} $? Let $a,b\in R$ and both are positive.
Can we say $\sqrt {ab}  \ge \min \{ a,b\} $?

Suppose without loss of generality that $0 < a \leq b$. As $a > 0$, we can multiply the last inequality by $a$. Then as $a^2 \leq ab$, we have $a \leq \sqrt{ab}$, which is what you are trying to prove.
