Prove that for any odd integer m, the number (m4 + 4m2 + 11)/16 is an integer.

We can express any odd integer $m$ in the form $m = 2k+1$ for some integer $k$.

Since $k$ and $k+1$ are consecutive integers, one of them must be even. Therefore, $k(k+1)$ is always an even integer. Let $k(k+1) = 2j$ for some integer $j$.
Then, $m^2 = 4(2j) + 1 = 8j+1$.
Now, find $m^4$:

The number is $\frac{m^4 + 4m^2 + 11}{16}$.
Substitute $m^4 = 64j^2 + 16j + 1$ and $m^2 = 8j+1$:

Factor out 16 from the numerator:

Since $j$ is an integer, $4j^2 + 3j + 1$ is also an integer.
Therefore, the given number is an integer.

Thus, $\sqrt{\frac{b}{a}} \times \sqrt{\frac{c}{b}} \times \sqrt{\frac{a}{c}} = 1$.

Substitute the given value of $\sqrt{n}$ into the expression:

For any real numbers $a$ and $b$, $(a-b)^2 \ge 0$.

We can divide the entire inequality by $ab$ without changing the direction of the inequality sign.

This proves the statement.

Since $x$ is a real number and $x \ne 0$, $x - \frac{1}{x}$ is a real number.
Therefore, $\left(x - \frac{1}{x}\right)^2 \ge 0$.

This proves the statement.