Integral Inequality
An inequality involving integrals of squares of functions.
Age
16 to 18
Challenge level
(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 $g(x)$, $$\left(\int_0^t
f(x)g(x)dx\right)^2 \leq \left(\int_0^t f(x)^2 dx\right)
\left(\int_0^t g(x)^2 dx\right).$$
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.
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: $$\left(\int_0^t f(x)g(x)\,dx\right)^2 \leq \left(\int_0^t f(x)^2\,dx\right) \left(\int_0^t g(x)^2\,dx\right).$$
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.