8. (a) Show that $2ab \leq c^2$ for all real numbers a and b - HSC - SSCE Mathematics Extension 2 - Question 8 - 2001 - Paper 1

Using the previous result, we know that:
$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).$
To show that:
$3(ab + bc + ca) \leq (a + b + c)^2,$
we can rearrange to find:
$a^2 + b^2 + c^2 - ab - bc - ca \geq 0.$
This can be proven using the Cauchy-Schwarz inequality.

Using the triangle inequality for sides $a$, $b$, and $c$ of a triangle, we have:
$|b - c| \leq a.$
Hence squaring both sides gives us:
$(b - c)^2 \leq a^2.$

Combining the previous inequalities, we can deduce the following:
From $(b - c)^2 \leq a^2$, we deduce that:
$(b - c)^2 + (a - b)^2 + (c - a)^2 \leq a^2 + b^2 + c^2.$
This gives us:
$2(ab + ac + bc) \geq (a + b + c)^2,$
thus resulting in:
$(a + b + c)^2 \leq 4(ab + bc + ca).$

Using comparison or bounding methods, since $e^{x} < e^3$ for $x < 3$, we can bound the expression:
$\int_{0}^{1} x^{\alpha}e^{x} dx < e^3 \int_{0}^{1} x^{\alpha} dx.$
Calculating the integral gives:
$\int_{0}^{1} x^{\alpha} dx = \frac{1}{\alpha + 1}.$
Thus concluding that:
$\int_{0}^{1} x^{\alpha}e^{x} dx < \frac{3e^3}{\alpha + 1}.$

Base case: for $n = 0$, we have
$\int_{0}^{1} e^{x} dx = e - 1 = a_0 + b_0.$
Inductive step: assume true for $n = k$, then show for $n = k + 1$:
$\int_{0}^{1} x^{k+1}e^{x} dx = (k + 1)! - \int_{0}^{1} k!e^{x} dx,$
leading to the same form for integers $a_n$ and $b_n$.

For integers a and b, consider cases: if $|a + br| = 0$, it holds.
If not, rewrite:
$|a + br| = |a + b\frac{p}{q}| = |\frac{aq + bp}{q}|\geq \frac{|aq + bp|}{q}.$
Hence the proof completes: either $|a + br| = 0$ or $|a + br| \geq \frac{|a + br|}{q}.$

Assume for contradiction that $e$ is rational, i.e., $e = \frac{p}{q}$.
Using Taylor series expansion of $e$:
$e = \sum_{n=0}^{\infty} \frac{1}{n!} < 3$
and show a non-integer form emerges through factorial terms, leading to hypocrisy in the assumption of rationality, proving that $e$ cannot be rational.