Prove the following inequality: $\int_{(a,b)}f\ d\lambda\cdot\int_{(a,b)}\frac{1}{f}d\lambda≥(b-a)^2$

Question

Assignment:

Let $-\infty < a < b < \infty$ and $f: (a,b) \rightarrow (0,\infty)$ be measurable, such that $f$ and $\frac{1}{f}$ are Lebesgue integrable. Prove the following inequality: $$\int_{(a,b)}f\ d\lambda\cdot\int_{(a,b)}\frac{1}{f}d\lambda≥(b-a)^2$$ Hint: Consider the integral $$\int_{(a,b)^2}\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)} \ d\lambda_2(x,y)$$

Thoughts: We have a theorem that says for an integrable sequence of functions $(f_n)_{n\in\mathbb{N}}$ that the new function $f(x_1,...,x_n) = f_1(x_1)\ \cdot...\cdot\ f_n(x_n)$ defined as the product of the single functions is integrable as well with the value of the integral of $f$ being the product of the integrals of $f_n$. This approach yields the equality, since, I think, I actually assume $x_1 = x_2$ and don't let the $x_i$ vary.

About the hint: The hint, I thought, suggests using Fubini, but since $f$ is random, I cannot simplify the integral at all. Also, using the theorem I referred to earlier, only yields, that the integral of the hint is the left side of the inequality times two, which doesn't help at all.

I'm stuck in a dead end right now, could anyone give me hint (or explain the hint given)?

Answer 1

Note

$$\int f\, d\lambda \cdot \int \frac{1}{f}\, d\lambda = \int_{(a,b)^2} \frac{f(x)}{f(y)} \, d\lambda_2 = \frac{1}{2} \int_{(a,b)^2} \frac{f(x)}{f(y)} + \frac{f(y)}{f(x)}\, d\lambda_2$$

and

$$\frac{f(x)}{f(y)} + \frac{f(y)}{f(x)} = \left(\sqrt{\frac{f(x)}{f(y)}} - \sqrt{\frac{f(y)}{f(x)}}\right)^2 + 2 \ge 2$$

on $(a,b)^2$.

Answer 2

As an alternative, note that this is the Cauchy-Schwarz inequality:

$$ (b-a)^2 = \left(\int_a^b \sqrt f\times \frac1{\sqrt f}\right)^2\le \int_a^b f\int_a^b\frac 1f $$