# Integral Inequality

An inequality involving integrals of squares of functions.

## Problem

(i)Suppose that $a$, $b$ and $t$ are positive. Which of
the following two expressions is the larger $$P=\left(\int_0^t
x^{a+b}dx\right)^2, \qquad Q=\left(\int_0^t x^{2a}dx \right)
\left(\int_0^t x^{2b}dx\right)\ ?$$ (ii)By considering
the inequality $$\int_0^t [f(x)+\lambda g(x)]^2 dx \geq 0,$$ prove
that, for all functions $f(x)$ and $g(x)$, $$\left(\int_0^t
f(x)g(x)dx\right)^2 \leq \left(\int_0^t f(x)^2 dx\right)
\left(\int_0^t g(x)^2 dx\right).$$

## Getting Started

The first part of this question involves an easy evaluation of the
integrals and then some simple algebra to compare the
results.

In the second part you will get a quadratic inequality in $\lambda$ of the form $$U + 2V\lambda + W\lambda^2 \geq 0.$$ Completing the square leads to a condition for this inequality to hold which gives the required result.

In the second part you will get a quadratic inequality in $\lambda$ of the form $$U + 2V\lambda + W\lambda^2 \geq 0.$$ Completing the square leads to a condition for this inequality to hold which gives the required result.

## Student Solutions

Shaun from Nottingham High School sent this solution.

(i) By calculation, we have $$P= \left({t^{a+b+1}\over a+b+1}\right)^2 = {t^{2(a+b+1)}\over (a+b+1)^2},$$ and $$Q= \left({t^{2a+1}\over 2a+1}\right)\left({t^{2b+1}\over 2b+1}\right) = {t^{2(a+b+1)}\over (2a+1)(2b+1)}.$$ Thus we must decide which of the two expressions $$(a+b+1)^2, \qquad (2a+1)(2b+1)$$ is the smaller. Since $$(a+b+1)^2-(2a+1)(2b+1) = (a-b)^2 \geq 0$$ we see that $P \leq Q$.

(ii) Given the inequality $$\int_0^t [f(x)+\lambda g(x)]^2\,dx \geq 0,$$ we have $$\lambda^2 \int_0^tg^2(x)dx+2\lambda\int_0^tg(x)f(x)dx +\int_0^tf^2(x)dx \geq 0$$ The discriminant of this quadratic in $\lambda$ must be less than or equal to 0 since the quadratic is never negative: $$4\left[\int_0^tg(x)f(x)dx\right]^2-4\int_0^tg^2(x)dx\int_0^tf^2(x)dx \leq 0.$$ This gives the required inequality which is known as the Cauchy Schwarz inequality: $$\left(\int_0^t f(x)g(x)\,dx\right)^2 \leq \left(\int_0^t f(x)^2\,dx\right) \left(\int_0^t g(x)^2\,dx\right).$$

## Teachers' Resources

Here the first part is a special case of the result in the second
part. However it is easier to prove the general case
directly.