Prove that $\tan\alpha\tan\beta\tan\gamma > 1$, where $\alpha$, $\beta$, $\gamma$ are angles in an acute triangle

Question

Needed to prove that

$$\tan \alpha \cdot \tan \beta \cdot \tan \gamma > 1$$ where $\alpha, \beta, \gamma < 90^\circ$ (parts of an acute triangle).

This question came up during a lesson and the teacher couldn't prove it with the methods we have.

Answer 1

We know that $\sum_{\alpha, \beta,\gamma} \tan \alpha=\prod_{\alpha, \beta,\gamma} \tan \alpha$ for angles of a triangle.

Now, all three angles of an acute angled triangle cannot be less than $45°$. Hence, $\sum_{\alpha,\beta,\gamma} \tan \alpha$ must be more than $1$, hence the product too.