8. Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 8 - 2001 - Paper 1

Using the result from part (i), we can expand $(a + b + c)^2$:

$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc).$

Thus:

$a^2 + b^2 + c^2 \geq 2ab \Rightarrow 3(ab + bc + ca) \leq a^2 + b^2 + c^2 + 2(ab + ac + bc) = (a + b + c)^2.$

For sides of a triangle, we can use the triangle inequality:

$a + b > c \Rightarrow b - c < a.$

Squaring both sides gives:

$(b - c)^2 < a^2.$

Therefore, we can conclude $(b - c)^2 \leq a^2$.

Using the results from previous parts, we can say:

Thus:

$(a + b + c)^2 \leq 4(ab + bc + ca).$

For positive $\alpha$, consider:

$\int_0^1 \alpha x e^x dx < \int_0^1 3 e^x dx.$

Evaluating, the right-hand side computes to:

$\frac{3}{\alpha + 1}$

Base case: For $n=0$, $\int_0^1 e^x dx = e - 1$. Assume true for $n=k$. Proceed to show for $k+1$ by integration by parts.

Since r is rational, if $|a + br| = 0$, the statement holds. If not, we can simplify the expression, $a + br = \frac{p}{q}$. Therefore, when $p$ and $q$ are positive integers, we derive:

$|a + br| \geq \frac{|a + br|}{q}.$

Assume to the contrary that e is rational, expressible as $\frac{p}{q}$. Consider the series:

$e = \sum_{n=0}^{\infty} \frac{1}{n!}.$

Show that no integer can achieve the equality for sufficiently large n, reaching a contradiction.