# Calculus inequality challenge prove ∫10f(x)/f(x+1/2)dx≥1

• MHB
• lfdahl
In summary, the calculus inequality challenge is a mathematical problem that involves proving the inequality ∫10f(x)/f(x+1/2)dx≥1. It is used to test understanding of key concepts in calculus such as integrals, functions, and inequalities. The notation ∫10f(x)/f(x+1/2)dx represents the integral of the function f(x)/f(x+1/2) over the interval [1,0], or the area under the curve between the x-values of 1 and 0. The significance of the inequality is that it shows the relationship between two functions and their areas under the curve. To prove the challenge, mathematical techniques such as integration, substitution, and algebraic
lfdahl
Gold Member
MHB
Let $f$ be a positive and continuous function on the real line which satisfies $f(x + 1) = f(x)$ for all numbers $x$.
Prove $\int_{0}^{1}\frac{f(x)}{f(x+\frac{1}{2})}dx \geq 1.$

Here is my solution.

Note $\int_0^{1/2} \frac{f(x)}{f(x + 1/2)}\, dx = \int_{1/2}^1 \frac{f(x + 1/2)}{f(x)}\, dx$ and $$\int_{1/2}^1 \frac{f(x)}{f(x + 1/2)}\, dx = \int_0^1 \frac{f(x + 1/2)}{f(x + 1)}\, dx = \int_0^{1/2} \frac{f(x + 1/2)}{f(x)}\, dx$$ Therefore
$$\int_0^1 \frac{f(x)}{f(x+1/2)}\, dx = \int_0^1 \frac{f(x+1/2)}{f(x)}\, dx$$ and consequently
$$\int_0^1 \frac{f(x)}{f(x+1/2)}\, dx = \frac{1}{2}\int_0^1 \left(\frac{f(x)}{f(x+1/2)} + \frac{f(x+1/2)}{f(x)}\right)\, dx \ge \frac{1}{2}\int_0^1 2\sqrt{\frac{f(x)}{f(x+1/2)}\cdot \frac{f(x+1/2)}{f(x)}}\, dx = 1$$

Euge said:
Here is my solution.

Note $\int_0^{1/2} \frac{f(x)}{f(x + 1/2)}\, dx = \int_{1/2}^1 \frac{f(x + 1/2)}{f(x)}\, dx$ and $$\int_{1/2}^1 \frac{f(x)}{f(x + 1/2)}\, dx = \int_0^1 \frac{f(x + 1/2)}{f(x + 1)}\, dx = \int_0^{1/2} \frac{f(x + 1/2)}{f(x)}\, dx$$ Therefore
$$\int_0^1 \frac{f(x)}{f(x+1/2)}\, dx = \int_0^1 \frac{f(x+1/2)}{f(x)}\, dx$$ and consequently
$$\int_0^1 \frac{f(x)}{f(x+1/2)}\, dx = \frac{1}{2}\int_0^1 \left(\frac{f(x)}{f(x+1/2)} + \frac{f(x+1/2)}{f(x)}\right)\, dx \ge \frac{1}{2}\int_0^1 2\sqrt{\frac{f(x)}{f(x+1/2)}\cdot \frac{f(x+1/2)}{f(x)}}\, dx = 1$$

Excellent, Euge! (Nod) Thankyou very much for your participation!

## What is calculus inequality challenge?

The calculus inequality challenge is a mathematical problem that involves proving the inequality: ∫10f(x)/f(x+1/2)dx≥1. It is a commonly used problem in calculus courses to test students' understanding of concepts such as integrals, functions, and inequalities.

## What does the notation ∫10f(x)/f(x+1/2)dx mean?

The notation ∫10f(x)/f(x+1/2)dx represents the integral of the function f(x)/f(x+1/2) over the interval [1,0]. This means that we are finding the area under the curve of the function f(x)/f(x+1/2) between the x-values of 1 and 0.

## What is the significance of the inequality in the calculus inequality challenge?

The inequality in the calculus inequality challenge is significant because it allows us to prove that the integral of a function divided by another function is greater than or equal to 1. This can be used to show the relationship between two functions and their areas under the curve.

## How do you prove the calculus inequality challenge?

To prove the calculus inequality challenge, we need to use mathematical techniques such as integration, substitution, and algebraic manipulation. We also need to have a strong understanding of the properties of integrals and inequalities.

## Why is the calculus inequality challenge important?

The calculus inequality challenge is important because it tests our understanding of key concepts in calculus and helps us develop problem-solving skills. It also has real-world applications in fields such as economics, physics, and engineering, where inequalities and integrals are used to model and solve problems.

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