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 smallest.

Since

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

we see that $P \leq Q$ .
(ii) By computing $[f (x) + λ g (x {)]}^{2}$ and integrating, we have

$U + 2 V λ + W λ^{2} \geq 0,$ $(⋆)$
for all $λ$ , where

$U = \int_{0}^{t} [f (x {)]}^{2} dx, V = \int_{0}^{t} f (x) g (x) dx, W = \int_{0}^{t} [g (x {)]}^{2} dx .$

The condition that $(⋆)$ holds for all $λ$ is $UW \geq V^{2}$ which is the required inequality.