Shaun from Nottingham High School sent this solution.

(i) By calculation, we have

P = {(\frac{t^{a + b + 1}}{a + b + 1})}^{2} = \frac{t^{2 (a + b + 1)}}{(a + b + 1)^{2}},

and

Q = (\frac{t^{2 a + 1}}{2 a + 1}) (\frac{t^{2 b + 1}}{2 b + 1}) = \frac{t^{2 (a + b + 1)}}{(2 a + 1) (2 b + 1)} .

Thus we must decide which of the two expressions

(a + b + 1)^{2}, (2 a + 1) (2 b + 1)

is the smaller. Since

(a + b + 1)^{2} - (2 a + 1) (2 b + 1) = (a - b)^{2} \geq 0

we see that

P \leq Q

.
(ii) Given the inequality

\int_{0}^{t} [f (x) + λ g (x {)]}^{2} dx \geq 0,

we have

λ^{2} \int_{0}^{t} g^{2} (x) dx + 2 λ \int_{0}^{t} g (x) f (x) dx + \int_{0}^{t} f^{2} (x) dx \geq 0

The discriminant of this quadratic in

λ

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 \leq 0 .

This gives the required inequality which is known as the Cauchy Schwarz inequality:

{(\int_{0}^{t} f (x) g (x) dx)}^{2} \leq (\int_{0}^{t} f (x)^{2} dx) (\int_{0}^{t} g (x)^{2} dx) .