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

Step 1: Express $m$ in terms of an integer $k$.
Since $m$ is an odd integer, we can write $m = 2k+1$ for some integer $k$.

Step 2: Calculate $m^2$.

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

Since $k$ and $k+1$ are consecutive integers, one of them must be even. Thus, $k(k+1)$ is always an even number. Let $k(k+1) = 2j$ for some integer $j$.

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

Step 3: Calculate $m^4$.

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

Step 4: Substitute $m^2$ and $m^4$ into the given expression.
The expression is $\frac{m^4 + 4m^2 + 11}{16}$.
Substitute the values:

$$\frac{(64j^2 + 16j + 1) + 4(8j+1) + 11}{16}$$ $$\frac{64j^2 + 16j + 1 + 32j + 4 + 11}{16}$$ $$\frac{64j^2 + (16j + 32j) + (1 + 4 + 11)}{16}$$ $$\frac{64j^2 + 48j + 16}{16}$$

Step 5: Simplify the expression.

$$\frac{64j^2}{16} + \frac{48j}{16} + \frac{16}{16}$$ $$4j^2 + 3j + 1$$

Since $j$ is an integer, $4j^2 + 3j + 1$ is also an integer.
Therefore, the number $\frac{m^4 + 4m^2 + 11}{16}$ is an integer.

The final answer is $\boxed{\text{The expression is an integer}}$.

Step 1: Combine the terms under a single square root.
Given $a, b, c$ are positive real numbers.
We use the property $\sqrt{x} \times \sqrt{y} = \sqrt{xy}$.

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

Step 2: Multiply the fractions inside the square root.

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

Step 3: Simplify the expression inside the square root.

$$\sqrt{\frac{abc}{abc}} = \sqrt{1}$$

Step 4: Evaluate the square root.

$$\sqrt{1} = 1$$

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

The final answer is $\boxed{1}$.

Step 1: Substitute the given value of $\sqrt{n}$.
We are given $\sqrt{n} = \frac{3}{2}$.
We need to find $-\sqrt{n}$.

Step 2: Calculate $-\sqrt{n}$.

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

The final answer is $\boxed{-\frac{3}{2}}$.

Step 1: Start with a known true inequality.
For any real numbers $a$ and $b$, the square of their difference is always non-negative:

$$(a-b)^2 \ge 0$$

Step 2: Expand the inequality.

$$a^2 - 2ab + b^2 \ge 0$$

Step 3: Add $2ab$ to both sides.

$$a^2 + b^2 \ge 2ab$$

Step 4: Divide both sides by $ab$.
Since $a$ and $b$ are positive real numbers, $ab$ is also positive. Dividing by a positive number does not change the direction of the inequality.

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

Step 5: Separate the terms on the left side and simplify.

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

This proves the inequality.

The final answer is $\boxed{\frac{a}{b} + \frac{b}{a} \ge 2}$.

Step 1: Start with a known true inequality.
For any real number $x$, the square of $(x - \frac{1}{x})$ is always non-negative, provided $x \ne 0$:

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