$\sin^2 \alpha + \sin^2 \beta - \cos \gamma < M$ given that the sum of the angles is $\pi$ 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?

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 $y=\pi$:
$$M=\max_x\ \sin^2(x)-\cos(x) = 5/4$$
