Inequalities Concerning Roots

Theorem

Let $\left[{X . . Y}\right]$ be a closed real interval such that $0 < X \le Y$.

Let $x, y \in \left[{X . . Y}\right]$.

Then:

$\forall n \in \N^*: X Y^{1/n} \left|{x - y}\right| \le n X Y \left|{x^{1/n} - y^{1/n}}\right| \le Y X^{1/n} \left|{x - y}\right|$

Proof

From Difference of Two Powers, we have that:

$\displaystyle a^n - b^n = \left({a - b}\right) \left({a^{n-1} + a^{n-2} b + a^{n-3} b^2 + \ldots + a b^{n-2} + b^{n-1}}\right) = \left({a - b}\right) \sum_{j=0}^{n-1} a^{n-j-1} b^j$

Let $a > b > 0$.

Then:

$\displaystyle \frac {a^n - b^n}{a - b} = a^{n-1} + a^{n-2} b + a^{n-3} b^2 + \ldots + a b^{n-2} + b^{n-1}$

Also, note that:

$b^{n-1} = b^{n-j-1} b^j \le a^{n-j-1} b^j \le a^{n-j-1} a^j \le a^{n-1}$

Hence if $a > b > 0$, we have:

$\displaystyle n b^{n-1} \le \frac {a^n - b^n}{a - b} \le n a^{n-1}$

Taking reciprocals, we get:

$\displaystyle \frac 1 {n b^{n-1}} \ge \frac {a - b}{a^n - b^n} \ge \frac 1 {n a^{n-1}}$

Thus:

$\displaystyle \frac {a^n - b^n} {n a^{n-1}} \le a - b \le \frac {a^n - b^n} {n b^{n-1}} \implies \frac {a \left({a^n - b^n}\right)} {a^n} \le n \left({a - b}\right) \le \frac {b \left({a^n - b^n}\right)} {b^n}$

Now suppose $x > y$. Then:

$\displaystyle x - y = \left|{x - y}\right|$ and $x^{1/n} - y^{1/n} = \left|{x^{1/n} - y^{1/n}}\right|$

Then put $a = x^{1/n}$ and $b = y^{1/n}$ in the above:

$\displaystyle \frac {x^{1/n} \left|{x - y}\right|} x \le n \left|{x^{1/n} - y^{1/n}}\right| \le \frac {y^{1/n} \left|{x - y}\right|} y$

Multiplying through by $x y$ gives us:

$x^{1/n} y \left|{x - y}\right| \le n x y \left|{x^{1/n} - y^{1/n}}\right| \le x y^{1/n} \left|{x - y}\right|$

Similarly, if $y > x$, we get:

$y^{1/n} x \left|{x - y}\right| \le n x y \left|{x^{1/n} - y^{1/n}}\right| \le y x^{1/n} \left|{x - y}\right|$

The result follows after some algebra.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 3.11 \ (6)$