Proof $\mid(a - b)^{\frac{1}{n}}\mid \leqslant \mid a^{\frac{1}{n}} - b^{\frac{1}{n}}\mid$

Question

We have $(a + b)^{n} \geq a^{n} + b^{n}$

Proof:

$\mid a - b \mid ^{\frac{1}{n}} \geq \mid \mid a \mid^{\frac{1}{n}} - \mid b\mid^{\frac{1}{n}}\mid$

Answer 1

You can show this by using induction.

https://en.wikipedia.org/wiki/Mathematical_induction#Description

I'll leave it to you to verify it using a base.

Now assume that

$\mid(a - b)^{\frac{1}{n}}\mid \leqslant \mid a^{\frac{1}{n}} - b^{\frac{1}{n}}\mid$ (1)

is true. We want to deduce that

$\mid(a - b)^{\frac{1}{n+1}}\mid \leqslant \mid a^{\frac{1}{n+1}} - b^{\frac{1}{n+1}}\mid$ (2)

is true as well.

Hint: Try and see if you can rewrite (2) to (1) in some way and use your assumption. Can you take it from here?

Answer 2

Raising to the $n$-th power, $|a - b|^{\frac{1}{n}} \geq || a|^{\frac{1}{n}} - |b|^{\frac{1}{n}}| $ becomes $|a - b| \geq || a|^{\frac{1}{n}} - |b|^{\frac{1}{n}}|^n $.

Letting $a = c^n$ and $b = d^n$, this becomes $|c^n - d^n| \geq || c| - |d||^n $.

Assuming that $|c| \ge |d|$, dividing by $c^n$ and letting $\frac{d}{c} = x$, this is $|1 - x^n| \geq |1 - x|^n $.

Assume that $1 \ge x \ge 0$. So we want to show that $1-x^n \ge (1-x)^n$.

This is true for $x=0$ and $x=1$.

This is true for $n=1$.

Since $x^n$ and $(1-x)^n$ are decreasing with increasing $n$, $1-x^n$ is increasing and $(1-x)^n$ is decreasing so the inequality holds for all $n$.