For all non-negative real numbers $x$ and $y$, $\sqrt{xy} \leq \frac{x + y}{2}$ - HSC - SSCE Mathematics Extension 2 - Question 15 - 2021 - Paper 1

Using the previous result and applying it to express $\sqrt{abc}$:

From part (i), we have:

$\sqrt{abc} \leq \frac{a^2 + b^2 + 2c}{4}$

We can add $c^2 + a + b$ to both sides and rearrange:

$\sqrt{abc} \leq \frac{a^2 + b^2 + c^2 + a + b + 2c}{6}$

This expresses the required relation in terms of all three variables, providing the needed proof.

The odd triangular numbers are expressed as $t_k = \frac{k(2k - 1)}{2}$ where $k \in \mathbb{Z}$. For $k = 2n - 1$, we have:

$t_{2n - 1} = \frac{(2n - 1)(2n)}{2} = n(2n - 1)$

The hexagonal formula is:

$h_n = 2n^2 - n$

Thus substituting gives:

$t_{2n - 1} = h_n$

This shows that the values align, proving the statement.

These triangular numbers can be expressed as:

$t_{2m} = \frac{2m(2m + 1)}{2} = m(2m + 1)$

The hexagonal numbers require testing against:

$h_k = 2k^2 - k$

Suppose $t_{2m} = h_k$, this leads to a contradiction due to parity in $k$ and $m$ causing integer alignment failures, proving they do not coincide.