Copyright © University of Cambridge. All rights reserved.

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

Show menu


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