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

Proof
From Difference of Two Powers, we have that $$a^n - b^n = \left({a - b}\right) \left({a^{n-1} + a^{n-2} b + a^{n-3} b^2 + \ldots + a b^{n-2} + b^{n-1}}\right) = \left({a - b}\right) \sum_{j=0}^{n-1} a^{n-j-1} b^j$$.

Let $$a > b > 0$$.

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

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

Taking reciprocals, we get $$\frac 1 {n b^{n-1}} \ge \frac {a - b}{a^n - b^n} \ge \frac 1 {n a^{n-1}}$$.

Thus $$\frac {a^n - b^n} {n a^{n-1}} \le a - b \le \frac {a^n - b^n} {n b^{n-1}} \Longrightarrow \frac {a \left({a^n - b^n}\right)} {a^n} \le n \left({a - b}\right) \le \frac {b \left({a^n - b^n}\right)} {b^n}$$.

Now suppose $$x > y$$. Then $$x - y = \left|{x - y}\right|$$ 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:

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

Multiplying through by $$xy$$ 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.