How do I prove this trigonometry inequality $\sin^2(y) \leq 2 \sin^2(x) + 2\sin^2(y-x)$?

by Cauchy Schwarz inequality $$2 (\sin^2 y+\sin^2 (y-x))\ge {\left(|\sin y|+|\sin(y-x)|\right)}^2$$ It remains to prove $$|\sin y|+|\sin(y-x)|\ge |\sin x|$$ which is true as by triangle inequality $$\begin{align}|\sin (x)|=|\sin (x-y+y)|=|\sin y\cos (x-y)+\sin (x-y)\cos y|&\\\le |\sin(y-x)|+|\sin y|\end{align}$$ We are done!

We used $|a+b|\le |a|+|b|$ and $|\cos (x-y)|\le 1 $ and $|\cos y|\le 1$

Let $x = a$, $y = a + b$. Calculate the difference between both sides.

\begin{align} & \sin^2(a+b) - 2\sin^2a - 2 \sin^2 b \\ &= (\sin a \cos b + \cos a \sin b)^2 - 2\sin^2 a - 2\sin^2 b \tag{compound-angle formula} \\ &= \sin^2 a (\cos^2 b - 1) + (\cos^2 a - 1) \sin^2 b - (\sin^2 a + \sin^2 b) + 2 \sin a \cos a \sin b \cos b \tag{split $\sin^2 a$ and $\sin^2 b$} \\ &\le -2\sin^2 a \sin^2 b - (\sin^2 a + \sin^2 b) + \frac12 (\sin^2 a + \sin^2 b) (\cos^2 a + \cos^2 b) \tag{AM-GM inequality} \\ &= -2\sin^2 a \sin^2 b - (\sin^2 a + \sin^2 b) + \frac12 (\sin^2 a + \sin^2 b) (2 - \sin^2 a - \sin^2 b) \\ &= 2\sin^2 a \sin^2 b - (\sin^2 a + \sin^2 b) + \frac12(\sin^2 a + \sin^2 b) (2) - \frac12(\sin^2 a + \sin^2 b)^2 \\ &= -2\sin^2 a \sin^2 b - \frac12 (\sin^2 a + \sin^2 b)^2 \\ &\le 0 \end{align}

Equality holds iff

1. $\sin a = \sin b$ and $\cos a = \cos b$ (from AM–GM inequality)
2. $\sin a \sin b = 0$ and $\sin^2 a + \sin^2 b = 0$ (from last step)

giving $\sin a = \sin b = 0$