Prove that $\tan \left ( \sum_{k=1}^{n} \theta_k \right ) \geq \sum_{k=1}^{n} \tan (\theta_k)$

Question

I'm trying to prove by induction that

$$\tan \left ( \sum_{k=1}^{n} \theta_k \right ) \geq \sum_{k=1}^{n} \tan (\theta_k)$$ provided that $$\sum_{k=1}^{n} \theta_k < \frac{\pi}{2}$$

So in the inductive step, we are thus required to prove that

$$\tan \left ( \sum_{k=1}^{n+1} \theta_k \right ) \geq \sum_{k=1}^{n+1} \tan (\theta_k)$$ or equivalently $$\tan \left ( \sum_{k=1}^{n+1} \theta_k \right ) - \sum_{k=1}^{n+1} \tan (\theta_k) \geq 0 $$

After some tinkering around and substituting in the inductive hypothesis, I end up showing that the expression on the left is greater than, or equal to

$$\sum_{k=1}^{n+1} \tan (\theta_k)\times \frac{ \tan (\theta_{n+1}) \tan \left ( \sum_{k=1}^{n} \theta_k \right )}{1- \tan (\theta_{n+1}) \tan \left ( \sum_{k=1}^{n} \theta_k \right ) }$$

Now I know that the sum on the left is positive, from the hypothesis that $$\sum_{k=1}^{n+1} \theta_k < \frac{\pi}{2}$$

So I now need to show that the fraction on the right is positive. But that does not seem to necessarily be the case. Did I go wrong somewhere here?

Answer 1

Note that $$\tan\left(\sum_{k=1}^{n+1}\theta_k\right)=\frac{\tan(\theta_{n+1})+\tan(\sum_{k=1}^n\theta_k)}{1-\tan(\theta_{n+1})\tan(\sum_{k=1}^n\theta_k)}$$

Since the LHS and the numerator are positive, the denominator must also be positive.

Answer 2

You need all angles to be positive angles! Then note that you have to show $\frac{1}{1-\tan(\theta_{n+1})\tan(\sum_{k=1}^{n}\theta_k)} \geq 1$ Which is equivalent to $1 \geq 1-\tan(\theta_{n+1})\tan(\sum_{k=1}^{n}\theta_k) $. Which follows from the positivity of the angles and the fact that $\sum_{k=1}^n \theta_k < \pi/2$ because ($\sum_{k=1}^{n+1} \theta_k < \pi/2$ and $\theta_k \geq 0$ hence).