# Prove that $\iint_{\left[ 0,1 \right] \times \left[ 0,1 \right]}{\frac{f\left( x \right)}{f\left( y \right)}\text{d}x\text{d}y\ge 1}$

Assume $$f(x)>0, x\in[0,1]$$. Prove that $$\iint_{\left[ 0,1 \right] \times \left[ 0,1 \right]}{\frac{f\left( x \right)}{f\left( y \right)}\text{d}x\text{d}y\ge 1}$$

It gives me the sense that the integration of $$\dfrac{f(x)}{f(y)}$$ will be small in one part of $$[0,1]\times [0,1]$$, but large in another part of $$[0,1]\times [0,1]$$ and finally their sum will be more than $$1$$. But how to formulate this sense formally? I guess we need to make some inequality scalings. Can anyone help?

• is $f$ continuous ?
– Surb
Dec 15 '20 at 12:36
• yeah $f$ is continuous Dec 15 '20 at 12:41
• 1. For $A>0$, $A + \frac{1}{A} \ge 2$; 2. $\iint \frac{f(x)}{f(y)} dx dy = \iint \frac{f(y)}{f(x)} dx dy$. Dec 15 '20 at 12:48

Suppose $$\int_{0}^{1} f$$ and $$\int_{0}^{1} \frac{1}{f}$$ exists and let $$I$$ be the given integral,
Observe that the given integral can be written as $$I = \left(\int_{0}^{1} f(x) \, dx \right)\left(\int_0^1\frac{1}{f(x)} \, dx\right)$$
Then by Cauchy Schwartz inequality for integrals we have $$I \geq \left(\int_0^1 1\, dx\right)^2 = 1$$