Let the real $x\in \left(0,\,\dfrac{\pi}{2}\right)$, prove that $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$.

Let the real $x\in \left(0,\,\dfrac{\pi}{2}\right)$, prove that $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$.

First, multiply the first and second fraction (top and bottom) from the LHS of the intended inequality by $\sin x$ and $\cos x$ respectively to get:

$\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}=\dfrac{\sin^4 x}{5\sin x}+\dfrac{\cos^4 x}{12 \cos x}$(*)

Since $\sin x$ and $\cos x$ are both positive reals in the given domain, we can apply the Titu's Lemma to the RHS of (*) to get:

$\begin{align*}\dfrac{\sin^4 x}{5\sin x}+\dfrac{\cos^4 x}{12 \cos x}&\ge \dfrac{(\sin^2 x+\cos^2 x)^2}{5\sin x+12 \cos x}\\&\ge \dfrac{1}{\sqrt{5^2+12^2}\sin \left(x+\tan^{-1}\dfrac{12}{5}\right)}\\&\ge \dfrac{1}{13 \sin \left(x+\tan^{-1}\dfrac{12}{5}\right)}\\&\ge \dfrac{1}{13}\,\,\,\text{since}\,\,\,\sin \left(x+\tan^{-1}\dfrac{12}{5}\right)\le 1 \,\,\,\text{for}\,\,\,  x\in \left(0,\,\dfrac{\pi}{2}\right)\end{align*}$