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Question: Find the least real value of $M$ such that the following inequality holds:

$$\sin^2 \alpha + \sin^2 \beta - \cos \gamma < M$$

Given that $\alpha, \beta, \gamma \in \mathbb{R}^+$, $\alpha + \beta + \gamma = \pi$

My attempt:

Step 1: Replace $\sin^2 t$ with $1 - \cos^2 t$

$2 - \cos^2 \alpha - \cos^2 \beta - \cos \gamma < M$

Furthermore, note that $- \cos \gamma = \cos (180 - \gamma) = \cos(\alpha + \beta)$

In addition, use this identity: $-\frac{1}{2}(\cos(2x) + \cos(2y)) = -\cos^2 x - \cos^2 y + 1$ to arrive at the following:

$$1 - \frac{1}{2}(\cos(2 \alpha) + \cos(2 \beta)) + \cos(\alpha + \beta) < M$$

And, conveniently, $\frac{1}{2} (\cos(2\alpha) + \cos(2\beta)) = \cos(\alpha + \beta) \cos(\alpha - \beta)$

$$1 - (\cos(\alpha + \beta))(\cos(\alpha - \beta) - 1) < M$$

$$ (\cos (\alpha + \beta))(1 - \cos(\alpha - \beta)) < M - 1$$

From the inequality $ab \leq \frac{(a + b)^2}{4}$, we have that $$(\cos (\alpha + \beta))(1 - \cos(\alpha - \beta)) \leq \frac{(1 + 2 \sin \alpha \sin \beta)^2}{4} < \frac{1}{4}$$

Since $0 < 2 \sin \alpha \sin \beta < 2$

That's all I have so far. Is it logical to then say that $M - 1 = \frac{1}{4}$? I don't think it is because that doesn't make sense to me (it's asking for the least M, and how do i know that $\frac{1}{4}$ is minimized?), but I am not very experienced by any means in dealing with inequalities. Though I do see that $\frac{5}{4}$ is approachable with $\alpha = \frac{\pi}{3} - h, \beta = h, \gamma = \frac{2 \pi}{3}$ where $h$ is an infitesimally small number.

Can anyone give me some guidance to finish up this question?

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1 Answer 1

up vote 1 down vote accepted

It should be a lot easier to look at the function: $$\sin^2(x)+ \sin^2(y)-\cos(\pi - x - y)$$ And note it is symmetric when interchanging $x$ and $y$, and noting that comparing it's derivatives to zero leads to $\sin(2x)=\sin(2y)$. Thus $x=y+n\pi$. Now find the maximum value of the function: $$\sin^2(x)+\sin^2(x+n \pi)-\cos(\pi-2 x-n \pi)$$ And show that $M=3$.

Edit:

As the OP's question constrains $x,y>0$, and since we have shown that there are no local maxima in the region, the maximum must lie on the boundary, i.e. either $x$ or $y$ must be either $0$ or $\pi$. Examine all four options and find for instance that when $y=\pi$: $$M=\max_x\ \sin^2(x)-\cos(x) = 5/4$$

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I am unfamiliar with this approach / unclear about what you're asking. May you explain further? –  Soke Feb 24 at 3:03
    
@user92774 - you are familiar with finding local minima / maxima by comparing the derivative to zero, right? –  nbubis Feb 24 at 3:34
    
Yes, but only with one variable. (I'm 80% through BC calculus irght now) –  Soke Feb 24 at 4:13
    
@user92774 - the same ideas hold for multiple variables - the derivative according to both $x$ and $y$ must be zero at the maximum point. –  nbubis Feb 24 at 4:34
    
Okay. That method gives me $M = 3$ but the argument doesn't satisfy the given domain (that the angles are positive reals) -- I get a maximum reached at $(\frac{\pi}{2}, \frac{-\pi}{2}, \pi)$ –  Soke Feb 24 at 4:42

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