Integral inequality

Shaun from Nottingham High School sent this solution.

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

(ii) Given the inequality $${\int }_0^t[f(x)+\lambda g(x){]}^{2}dx\ge 0,$$ we have $${\lambda }^2{\int }_0^t{g}^2(x)dx+2\lambda {\int }_0^{t}g(x)f(x)dx+{\int }_0^t{f}^2(x)dx\ge 0$$ The discriminant of this quadratic in $\lambda$ must be less than or equal to 0 since the quadratic is never negative: $$4{[{\int }_0^{t}g(x)f(x)dx]}^2-4{\int }_0^t{g}^2(x)dx{\int }_0^t{f}^2(x)dx\le 0.$$ This gives the required inequality which is known as the Cauchy Schwarz inequality: $${({\int }_0^{t}f(x)g(x)dx)}^2\le ({\int }_0^{t}f(x{)}^{2}dx)({\int }_0^{t}g(x{)}^{2}dx).$$