Suppose $f$ is the probability density function of a continuous random variable, and that $f^{\prime\prime}$ exists. I am trying to prove $$\left|\int_{-\infty}^{\infty}(f^2f^{\prime\prime})^{2/5}(x)\text{ d}x \right| \leq \int_{-\infty}^{\infty}\left|f^2f^{\prime\prime}\right|^{2/5}(x)\text{ d}x\text{.}$$ My understanding is that this is a consequence of Jensen's Inequality, but I'm not confident I understand why.
If I were to rewrite this in probabilistic terms, I suppose I could rewrite this as $$\left|\int_{-\infty}^{\infty}\dfrac{f^{\prime\prime (2/5)}(x)}{f^{1/5}(x)}f(x)\text{ d}x \right| \leq \int_{-\infty}^{\infty}\left|\dfrac{f^{\prime\prime (2/5)}(x)}{f^{1/5}(x)}\right|f(x)\text{ d}x\text{.}$$ so this would be the same thing as saying the following: $$\varphi\left(\mathbb{E}\left[\dfrac{f^{\prime\prime(2/5)}(X)}{f^{1/5}(X)} \right] \right) \leq \mathbb{E}\left[\varphi\left(\dfrac{f^{\prime\prime(2/5)}(X)}{f^{1/5}(X)} \right)\right]$$ where $\varphi(x) = |x|$. This should be true because $\varphi$ is convex.
Is my method correct? Or is it too complicated of a way to approach this problem?
What you have written is correct but the inequality $|\int g(X) d\mu (x)| \leq \int |g(x)| d\mu(x)$ is much more elementary than Jensen's inequality and convexity of $x \to |x|$. you can prove it by just writing $g$ as $g^{+}-g^{-}$.