Can we say $\sqrt {ab} \ge \min \{ a,b\} $?

Question

Let $a,b\in R$ and both are positive.

Can we say $\sqrt {ab} \ge \min \{ a,b\} $?

Answer 1

Hint: $\sqrt{ab}\geq\sqrt{\min\{a,b\}\min\{a,b\}}=\cdots$

Answer 2

Certainly. The square root function is monotonic, so (assuming WOLOG $a \ge b$) $\sqrt{ab}=\sqrt a \sqrt b \ge \sqrt b \sqrt b =b = \min \{a,b\}$

Answer 3

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.

Answer 4

The answer is yes. Moreover, $ \min\{a,b\} = \sqrt{a b} \iff a=b$.

If it helps, you might take a look at https://en.wikipedia.org/wiki/Generalized_mean; in fact

$$ \min\{a, b\} = M_{- \infty}(a,b) \leq M_p(a,b) = \left( \frac{a^p + b^p}{2}\right) ^ {\frac{1}{p}} \leq M_0(a,b) = \sqrt{a b}, \quad \forall p<0. $$

The special case in which $p=-1$ gives the so-called harmonic mean.