Q.The ratio of the A.M. and G.M. of two positive numbers $a$ and $b$, is $m : n$. Show that $a : b = \left(m + \sqrt{m^2 - n^2}\right) : \left(m - \sqrt{m^2 - n^2}\right)$.

Concept & Intuition

We have two positive numbers: $5^x$ and $5^{1-x}$. Their sum is what we want to minimize. For positive numbers, the arithmetic mean is always at least the geometric mean, with equality when the numbers are equal. That gives a lower bound on the sum — and that bound is actually achievable, so it’s the minimum.

1. Set up the AM–GM inequality For any positive $a$ and $b$,

Here $a = 5^x$, $b = 5^{1-x}$. So

2. Simplify the geometric mean

Hence

3. Apply the inequality

4. Check when equality occurs AM = GM when $a = b$, i.e. $5^x = 5^{1-x} \quad\Rightarrow\quad x = 1-x \quad\Rightarrow\quad x = \frac12.$ …

We start with the given equation:

for real values of $x$. The question asks what we can conclude about $\theta$.

Concept and Intuition

The expression $x + \frac{1}{x}$ is famous for having a restricted range when $x$ is real. Why? Because if $x > 0$, by AM–GM inequality, $x + \frac{1}{x} \ge 2$, with equality only at $x = 1$. If $x < 0$, then let $x = -t$ with $t > 0$, so $x + \frac{1}{x} = -t - \frac{1}{t} = -\left(t + \frac{1}{t}\right) \le -2$. So the sum is always either $\ge 2$ or $\le -2$.

Meanwhile, $\cos\theta$ for any real $\theta$ is always between $-1$ and $1$ inclusive. So we are equating a number that lives outside $[-1, 1]$ (except possibly at the boundaries) to a number that must live inside $[-1, 1]$. The only chance would be if $x + \frac{1}{x}$ could equal something in $[-1, 1]$, but it cannot — the gap is clear.

Thus, no real $\theta$ can satisfy the equation for any real $x$.

Step-by-step reasoning

1. Recall the range of $x + \frac{1}{x}$ for real $x$ For $x > 0$, by AM–GM:

For $x < 0$, set $x = -t$ with $t > 0$:

For $x = 0$, the expression is undefined. So the possible values are $(-\infty, -2] \cup [2, \infty)$.

2. Recall the range of $\cos\theta$ for real $\theta$ The cosine function always satisfies:

3. Compare the two ranges …

We are given two positive numbers in the ratio

We need the ratio of their arithmetic mean (AM) to their geometric mean (GM).

Concept and intuition:

When two numbers are in a given ratio, we can represent them as multiples of that ratio. The AM and GM are symmetric functions of the numbers, so the ratio of AM to GM will be independent of the scaling factor. The trick is to notice that $(3+2\sqrt{2})$ and $(3-2\sqrt{2})$ are conjugates — their product is a perfect square, which simplifies the GM nicely.

Step-by-step solution:

1. Set up the numbers. Let the two numbers be

where $k > 0$ is a common factor. This preserves the given ratio.

2. Compute the arithmetic mean (AM).

The $2\sqrt{2}$ terms cancel:

3. Compute the geometric mean (GM).

The product inside is

Using the difference of squares:

$(3)^2 - (2\sqrt{2})^2 = 9 - 8 = 1.$ …