Proving Inequalities: Squares, Surds and Telescoping Sums

Grades 10-11

Interactive Proving Inequalities: Squares, Surds and Telescoping Sums graph — click to explore — Click the graph to explore it interactively →

Question: Q6

To solve Question 6, we apply properties of quadratic functions, algebraic manipulation of square roots (rationalization), and the concept of telescoping sums.

Part (a): Squares and Positive Numbers

For any two real numbers

x

and

y

, we can factor the difference of their squares:

y^2 - x^2 = (y - x)(y + x)

x

and

y

are positive, then their sum

(y + x)

must be strictly greater than zero. Consequently, the sign of the expression

y^2 - x^2

is entirely determined by the sign of

(y - x)

If $x < y$ , then $(y - x) > 0$ , so $y^2 - x^2 > 0$ , which implies $x^2 < y^2$ .
If $x^2 < y^2$ , then $(y^2 - x^2) > 0$ , so $(y - x)(y + x) > 0$ . Since $(y + x) > 0$ , it follows that $(y - x) > 0$ , meaning $x < y$ .

This result is fundamental because it allows us to square both sides of an inequality (or take the square root) involving positive numbers without changing the direction of the inequality.

Part (b): Square Root Inequalities

(i) Proving $\sqrt{k+1} + \sqrt{k-1} < 2\sqrt{k}$ Since both sides are positive, we can square them using the result from part (a):

\begin{align*} (\sqrt{k+1} + \sqrt{k-1})^2 &= (k+1) + (k-1) + 2\sqrt{(k+1)(k-1)} \\ &= 2k + 2\sqrt{k^2 - 1} \end{align*}

The square of the right-hand side is

(2\sqrt{k})^2 = 4k = 2k + 2k

. Comparing the two, since

k^2 - 1 < k^2

for any positive integer

k

, we know

\sqrt{k^2 - 1} < k

. Thus,

2k + 2\sqrt{k^2 - 1} < 2k + 2k = 4k

. Taking the square root of both sides proves the inequality.

Geometric Intuition: This inequality arises because the function

f(x) = \sqrt{x}

is concave. In the diagram, the blue curve representing

\sqrt{x}

always stays above any chord (the red segment) connecting two points on the curve. The midpoint of the chord represents the average height

\frac{\sqrt{k+1} + \sqrt{k-1}}{2}

, which is visually lower than the curve height

\sqrt{k}

at the midpoint

x=k

(ii) Proving $\frac{1}{\sqrt{k}} < \sqrt{k+1} - \sqrt{k-1}$ Starting from the result in (i):

\sqrt{k+1} + \sqrt{k-1} < 2\sqrt{k}

Since

(\sqrt{k+1} - \sqrt{k-1})

is positive, we multiply both sides by it:

(\sqrt{k+1} + \sqrt{k-1})(\sqrt{k+1} - \sqrt{k-1}) < 2\sqrt{k}(\sqrt{k+1} - \sqrt{k-1})

(k+1) - (k-1) < 2\sqrt{k}(\sqrt{k+1} - \sqrt{k-1})

2 < 2\sqrt{k}(\sqrt{k+1} - \sqrt{k-1})

Dividing by

2\sqrt{k}

\frac{1}{\sqrt{k}} < \sqrt{k+1} - \sqrt{k-1}

Part (c): The Telescoping Sum

To prove

\sum_{k=1}^n \frac{1}{\sqrt{k}} < \sqrt{n+1} + \sqrt{n} - 1

, we substitute the inequality from (b)(ii) into each term of the sum:

\begin{align*} \sum_{k=1}^n \frac{1}{\sqrt{k}} &< \sum_{k=1}^n (\sqrt{k+1} - \sqrt{k-1}) \\ &< (\sqrt{2} - \sqrt{0}) + (\sqrt{3} - \sqrt{1}) + (\sqrt{4} - \sqrt{2}) + \dots + (\sqrt{n} - \sqrt{n-2}) + (\sqrt{n+1} - \sqrt{n-1}) \end{align*}

This is a telescoping sum. Notice how

\sqrt{2}

cancels with

-\sqrt{2}

\sqrt{3}

will cancel with a later term, and so on. Only the terms that do not have a pair remain:

The negative terms at the start: $-\sqrt{0}$ and $-\sqrt{1}$ .
The positive terms at the end: $\sqrt{n}$ and $\sqrt{n+1}$ . Thus, the sum simplifies to:

\sum_{k=1}^n \frac{1}{\sqrt{k}} < \sqrt{n+1} + \sqrt{n} - 1

The interactive diagram allows you to change the values of

k

and

n

to see these relationships hold true numerically and geometrically.