Assume $x \geq 1$ and deduce $x^{1/n} \geq 1$

Question

What I've done:

Assume$$x^{1/n}<1$$ $$(x^{1/n})^n<1^{n}$$ $$x<1$$ So we have a contradiction. Is this sufficient?

Answer 1

Your proof is valid.

You could have a direct proof as well

Note that the function $f(x)= x^{1/n}$ is strictly increasing on $(0,\infty)$ since its derivative is positive.

Therefore, $$ x>1\implies x^{1/n}>1^{1/n}=1$$

Answer 2

$x\ge1\implies\ln x\ge0\implies\dfrac1n\ln x\ge0\implies\ln x^{\frac1n}\ge0\implies x^{\frac1n}\ge e^0=1$

Answer 3

$$x^{1/n}-1=\frac{x-1}{x^{(n-1)/n}+\cdots+x^{1/n}+1}$$

Answer 4

Yes your proof is valid, indeed since $f(x)=x^\frac1n$ for $x\ge 1$ is an increasing function we have

$$x\ge 1 \iff x^{1/n}\ge 1$$