Problem: If [imath]p[/imath] is the first of [imath]n[/imath] arithmetic means between two numbers, and [imath]q[/imath] is the first of [imath]n[/imath] harmonic means between the same two numbers, prove that the value of [imath]q[/imath] cannot lie between [imath]p[/imath] and [imath]\left(\frac{n+1}{n-1}\right)^2 p[/imath].
Solution (written for feedback)
Let [imath]a < b[/imath] be the two positive numbers. If [imath]n[/imath] arithmetic means are inserted between [imath]a[/imath] and [imath]b[/imath], the first arithmetic mean [imath]p[/imath] is given by
[math]p = a + \frac{b-a}{n+1},[/math]which can be rearranged as
[math](n+1)p = na + b.[/math]
Similarly, if [imath]n[/imath] harmonic means are inserted between [imath]a[/imath] and [imath]b[/imath], the first harmonic mean [imath]q[/imath] satisfies
[math]\frac{1}{q} = \frac{1}{a} + \frac{\frac{1}{b}-\frac{1}{a}}{n+1}.[/math]After simplification, this becomes
[math]\frac{ab(n+1)}{q} = nb + a.[/math]
To eliminate [imath]b[/imath], we solve the arithmetic mean formula for [imath]b[/imath]:
[math]b = (n+1)p - na.[/math]Substituting this into the harmonic mean equation yields
[math]\frac{a(n+1)((n+1)p - na)}{q} = n((n+1)p - na) + a,[/math]which simplifies to the quadratic equation
[math]na^2 - a\big((n+1)p + (n-1)q\big) + npq = 0.[/math]
Since [imath]a[/imath] is a real number, the discriminant of this quadratic must be non-negative. Therefore,
[math]\Delta = \big((n+1)p + (n-1)q\big)^2 - 4 n^2 pq \ge 0.[/math]Expanding and simplifying gives
[math](n+1)^2 p^2 - 2 pq (n^2 + 1) + (n-1)^2 q^2 \ge 0,[/math]which factorises neatly as
[math](p(n+1)^2 - q(n-1)^2)(p - q) \ge 0.[/math]
This inequality gives two possible cases. In the first case, if both factors are non-negative,
[math]p(n+1)^2 - q(n-1)^2 \ge 0 \quad \text{and} \quad p - q \ge 0,[/math]then it follows that
[math]p(n+1)^2 \ge q(n-1)^2 \quad \text{and} \quad p \ge q.[/math]Multiplying [imath]p \ge q[/imath] by [imath](n-1)^2 > 0[/imath] and combining with the first inequality yields
[math]\frac{p(n+1)^2}{(n-1)^2} \ge p \ge q.[/math]
In the second case, if both factors are non-positive,
[math]p(n+1)^2 - q(n-1)^2 \le 0 \quad \text{and} \quad p - q \le 0,[/math]then we have
[math]q \ge \frac{p(n+1)^2}{(n-1)^2} \quad \text{and} \quad q \ge p,[/math]so that
[math]q \ge \frac{p(n+1)^2}{(n-1)^2} \ge p.[/math]
In either case, it is clear that [imath]q[/imath] cannot lie strictly between [imath]p[/imath] and [imath]\left(\frac{n+1}{n-1}\right)^2 p[/imath]. Therefore, we conclude
[math]q \notin \left( p, \left(\frac{n+1}{n-1}\right)^2 p \right).[/math]
Solution (written for feedback)
Let [imath]a < b[/imath] be the two positive numbers. If [imath]n[/imath] arithmetic means are inserted between [imath]a[/imath] and [imath]b[/imath], the first arithmetic mean [imath]p[/imath] is given by
[math]p = a + \frac{b-a}{n+1},[/math]which can be rearranged as
[math](n+1)p = na + b.[/math]
Similarly, if [imath]n[/imath] harmonic means are inserted between [imath]a[/imath] and [imath]b[/imath], the first harmonic mean [imath]q[/imath] satisfies
[math]\frac{1}{q} = \frac{1}{a} + \frac{\frac{1}{b}-\frac{1}{a}}{n+1}.[/math]After simplification, this becomes
[math]\frac{ab(n+1)}{q} = nb + a.[/math]
To eliminate [imath]b[/imath], we solve the arithmetic mean formula for [imath]b[/imath]:
[math]b = (n+1)p - na.[/math]Substituting this into the harmonic mean equation yields
[math]\frac{a(n+1)((n+1)p - na)}{q} = n((n+1)p - na) + a,[/math]which simplifies to the quadratic equation
[math]na^2 - a\big((n+1)p + (n-1)q\big) + npq = 0.[/math]
Since [imath]a[/imath] is a real number, the discriminant of this quadratic must be non-negative. Therefore,
[math]\Delta = \big((n+1)p + (n-1)q\big)^2 - 4 n^2 pq \ge 0.[/math]Expanding and simplifying gives
[math](n+1)^2 p^2 - 2 pq (n^2 + 1) + (n-1)^2 q^2 \ge 0,[/math]which factorises neatly as
[math](p(n+1)^2 - q(n-1)^2)(p - q) \ge 0.[/math]
This inequality gives two possible cases. In the first case, if both factors are non-negative,
[math]p(n+1)^2 - q(n-1)^2 \ge 0 \quad \text{and} \quad p - q \ge 0,[/math]then it follows that
[math]p(n+1)^2 \ge q(n-1)^2 \quad \text{and} \quad p \ge q.[/math]Multiplying [imath]p \ge q[/imath] by [imath](n-1)^2 > 0[/imath] and combining with the first inequality yields
[math]\frac{p(n+1)^2}{(n-1)^2} \ge p \ge q.[/math]
In the second case, if both factors are non-positive,
[math]p(n+1)^2 - q(n-1)^2 \le 0 \quad \text{and} \quad p - q \le 0,[/math]then we have
[math]q \ge \frac{p(n+1)^2}{(n-1)^2} \quad \text{and} \quad q \ge p,[/math]so that
[math]q \ge \frac{p(n+1)^2}{(n-1)^2} \ge p.[/math]
In either case, it is clear that [imath]q[/imath] cannot lie strictly between [imath]p[/imath] and [imath]\left(\frac{n+1}{n-1}\right)^2 p[/imath]. Therefore, we conclude
[math]q \notin \left( p, \left(\frac{n+1}{n-1}\right)^2 p \right).[/math]