Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $0<a<\pi/2,0<b<\pi/2$, $0<\lambda<1, \mu=1-\lambda$. Does anyone see a good proof of the inequality:

$$\sin(\lambda a)\sin(\lambda b)+\sin(\lambda a)\sin(\mu b)\cos(b)+\sin(\lambda b)\sin(\mu a)\cos(a)+\sin(\mu a)\sin(\mu b)>\sin(a)\sin(b).$$

share|cite|improve this question
    
The LHS equals the RHS for $a=b=\pi/4,\lambda=\mu=1/2$. – mathlove Dec 19 '15 at 21:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.