Question: By using Holder's inequality, prove that for any real number $\lambda,$ $$4\sin^2\lambda - \lambda \sin(2\lambda)\leq 2\lambda^2.$$
Assume that $\lambda\geq 0.$ Since Holder's inequality involves integral, I try to obtain $f(x)$ such that $$\int_0^\lambda f(x)\,dx = 4\sin^2(\lambda) - \lambda \sin(2\lambda).$$ By using first fundamental theorem of calculus and Wolfram, I obtain that $$f(\lambda)= 3\sin(2\lambda) - 2\lambda \cos(2\lambda).$$ Therefore, by Holder's inequality, \begin{align} \int_0^\lambda 3\sin(2x) - 2x \cos(2x)\,dx & \leq \left( \int_0^\lambda (3\sin(2x) - 2x \cos(2x))^2 \right)^{\frac{1}{2}} \left(\int_0^\lambda 1^2 \, dx\right)^{\frac{1}{2}}\\ & =\sqrt{\lambda}\left( \int_0^\lambda 9\sin^2(2x) - 12x\sin(2x)\cos(2x) + 4x^2\cos^2(2x) \right)^{\frac{1}{2}}\\ & \leq \sqrt{\lambda} \left( \int_0^\lambda 9 +6x+4x^2\,dx \right)^{\frac{1}{2}} \\ & = \sqrt{\lambda} \left(9\lambda + 3\lambda^2+ \frac{4}{3}\lambda^3\right)^{\frac{1}{2}}. \end{align} However, it does not seem like the inequality $$9\lambda^2 + 3\lambda^3+\frac{4}{3}\lambda^4 \leq 4\lambda^4$$ holds. The following is their graphs.
So I run out of idea to prove the inequality. Any hint would be appreciated.
Let \begin{equation} f(x) = \frac{3\sin(2x) - 2x \cos(2x)}{4x},\quad \text{and} \quad g(x)=4x. \end{equation} We apply Hölder's inequality on the interval $[0,\lambda]$ for $\lambda > 0$: \begin{equation} \int_0^\lambda f(x)g(x)\,dx \le ||fg||_1 \le ||f||_\infty ||g||_1. \end{equation} The supremum of $|f|$ over $\mathbb{R}$ is $1$, and $||g||_1 = 2 \lambda^2$, so \begin{equation} 4 \sin^2 \lambda - \lambda \sin(2\lambda) \le 2 \lambda^2 \end{equation} for all $\lambda > 0$. The result is trivial for $\lambda = 0$, and the functions in the inequality are even, so the inequality holds for all real $\lambda$.