If m is an odd integer, prove that (m4 + 4m2 + 11) / 16 is an integer. If a, b, c are positive real numbers, prove that sqrt(b/a) sqrt(c/b) sqrt(a/c) = 1.

Let $m$ be an odd integer. We need to show that the number $\frac{m^4 + 4m^2 + 11}{16}$ is an integer.
Since $m$ is an odd integer, we can write $m = 2k+1$ for some integer $k$.

First, let's find $m^2$:

$$m^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$$

The product of two consecutive integers, $k(k+1)$, is always an even number. So, $k(k+1) = 2j$ for some integer $j$.
Therefore, $m^2 = 4(2j) + 1 = 8j + 1$. This means $m^2 \equiv 1 \pmod{8}$.

Next, let's find $m^4$:

$$m^4 = (m^2)^2 = (8j+1)^2 = 64j^2 + 16j + 1 = 16(4j^2 + j) + 1$$

This means $m^4 \equiv 1 \pmod{16}$. Let $p = 4j^2 + j$, which is an integer. So, $m^4 = 16p + 1$.

Now substitute $m^4 = 16p+1$ and $m^2 = 8j+1$ into the given expression:

$$\frac{m^4 + 4m^2 + 11}{16} = \frac{(16p+1) + 4(8j+1) + 11}{16}$$ $$= \frac{16p + 1 + 32j + 4 + 11}{16}$$ $$= \frac{16p + 32j + 16}{16}$$

Factor out 16 from the numerator:

$$= \frac{16(p + 2j + 1)}{16}$$ $$= p + 2j + 1$$

Since $p$ and $j$ are integers, their sum $p+2j+1$ is also an integer.
Thus, the number $\frac{m^4 + 4m^2 + 11}{16}$ is an integer.
$\boxed{\text{The expression simplifies to an integer } p+2j+1.}$

Given that $a, b, c$ are positive real numbers, we need to show that $\sqrt{\frac{b}{a}} \times \sqrt{\frac{c}{b}} \times \sqrt{\frac{a}{c}} = 1$.
We can use the property of square roots that $\sqrt{x} \times \sqrt{y} = \sqrt{xy}$ and $\sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}}$.
Combine the terms under a single square root:

$$\sqrt{\frac{b}{a}} \times \sqrt{\frac{c}{b}} \times \sqrt{\frac{a}{c}} = \sqrt{\frac{b}{a} \times \frac{c}{b} \times \frac{a}{c}}$$

Multiply the fractions inside the square root:

$$= \sqrt{\frac{b \cdot c \cdot a}{a \cdot b \cdot c}}$$

Cancel out the common terms in the numerator and denominator:

$$= \sqrt{1}$$ $$= 1$$

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

Given $\sqrt{n} = \frac{3}{2}$. We need to find $-\sqrt{n}$.
Substitute the given value of $\sqrt{n}$ into the expression $-\sqrt{n}$:

$$-\sqrt{n} = -\left(\frac{3}{2}\right)$$ $$-\sqrt{n} = -\frac{3}{2}$$

$\boxed{-\frac{3}{2}}$

Let $a$ and $b$ be positive real numbers. We need to show that $\frac{a}{b} + \frac{b}{a} \ge 2$.
We know that for any real number $x$, $(x-y)^2 \ge 0$.
Consider the expression $\left(\sqrt{\frac{a}{b}} - \sqrt{\frac{b}{a}}\right)^2$. Since $a$ and $b$ are positive, $\sqrt{\frac{a}{b}}$ and $\sqrt{\frac{b}{a}}$ are real numbers.
Therefore,

$$\left(\sqrt{\frac{a}{b}} - \sqrt{\frac{b}{a}}\right)^2 \ge 0$$

Expand the square:

$$\left(\sqrt{\frac{a}{b}}\right)^2 - 2\sqrt{\frac{a}{b}}\sqrt{\frac{b}{a}} + \left(\sqrt{\frac{b}{a}}\right)^2 \ge 0$$ $$\frac{a}{b} - 2\sqrt{\frac{a}{b} \cdot \frac{b}{a}} + \frac{b}{a} \ge 0$$ $$\frac{a}{b} - 2\sqrt{1} + \frac{b}{a} \ge 0$$ $$\frac{a}{b} - 2 + \frac{b}{a} \ge 0$$

Add 2 to both sides of the inequality:

$$\frac{a}{b} + \frac{b}{a} \ge 2$$

This proves the statement.
$\boxed{\frac{a}{b} + \frac{b}{a} \ge 2}$

Let $x$ be a real number. We need to show that $x^2 + \frac{1}{x^2} \ge 2$, for $x \ne 0$.
We know that the square of any real number is non-negative.
Consider the expression $\left(x - \frac{1}{x}\right)^2$. Since $x$ is a real number and $x \ne 0$, $x - \frac{1}{x}$ is a real number.
Therefore,

$$\left(x - \frac{1}{x}\right)^2 \ge 0$$

Expand the square:

$$x^2 - 2 \cdot x \cdot \frac{1}{x} + \left(\frac{1}{x}\right)^2 \ge 0$$ $$x^2 - 2 + \frac{1}{x^2} \ge 0$$

Add 2 to both sides of the inequality:

$$x^2 + \frac{1}{x^2} \ge 2$$

This proves the statement.
$\boxed{x^2 + \frac{1}{x^2} \ge 2}$