Copyright © University of Cambridge. All rights reserved.

## 'Integral Inequality' printed from http://nrich.maths.org/

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: $$\left(\int_0^t f(x)g(x)\,dx\right)^2
\leq \left(\int_0^t f(x)^2\,dx\right) \left(\int_0^t
g(x)^2\,dx\right).$$