An algebra problem by Aly Ahmed

$\cos (α-β)\leq 1\implies 1-\cos α\cos β\geq \sin α\sin β\implies 2\sin α(1-\cos α\cos β)\geq 2 \sin^2 α\sin β\implies 2\sin α-\sin β\geq \sin (2α-β)$

Substituting $α$ by $2x$ and $β$ by $2y$ we get $2\sin 2x-\sin 2y\geq \sin (4x-2y)$

$\implies 3(\sin 2x+\sin 2y)-\sin (4x-2y)-\sin 2x\geq 4\sin 2y$

$\implies \cos (x-y) \left (3\sin (x+y)-\sin (3x-y)\right )\geq 2\sin 2y$

$\implies 4\cos (x-y) (\sin y\cos^3 x+\cos y\sin^3 x) \geq 2\sin 2y$

$\implies \cos (x-y) \left (\dfrac {\cos^3 x}{\cos y}+\dfrac {\sin^3 x}{\sin y}\right )\geq 1$

$\implies \dfrac {\cos^3 x}{\cos y}+\dfrac {\sin^3 x}{\sin y}\geq \sec (x-y)$

$\implies \dfrac {\cos^3 x}{\cos y}+\dfrac {\sin^3 x}{\sin y}-\sec (x-y) \geq \boxed 0$ .