Prove that $x < y$ implies $x^{1/n} < y^{1/n}$. Let $x$ and $y$ be positive real numbers and let $m$ and $n$ be positive integers.
Prove that $x < y$ implies $x^{1/n} < y^{1/n}$.
I have so far figured out two different approaches to prove this.

*

*By induction:

Base case: $P(1)\iff x^{1/1} < y^{1/1} \iff x<y$
Inductive step: $P(n+1)\iff x^{1/n+1} < y^{1/n+1} $
But then I'm stuck from here as I 'm not sure on how to solve $1/n+1$.


*Another approach is to use direct proof:

By rules of inequalities we know
$x<y\implies x-x < y-x$
$\implies xy < yy$
$\implies xx < xy$
Hence $x^{1/n} < y^{1/n}$
as
$(x^1)^{1/n} < (y^1)^{1/n}$
May I ask how can I solve for $1/n+1$ at the first approach and is the second approach correct at all?
 A: One can argue this way: Let if possible, there exists some $n$ such that $x^{\frac{1}{n}}\geqslant y^{\frac{1}{n}}$ then take nth power both sides and we get $x\geqslant y$. Everything is fine with inequality as $x$ and $y$ are positive.
A: First we prove a simple property. Given $a,b,c,d$ are positive real numbers
(eqn 1)
$a \ge c$ and $b \ge d \implies ab \ge cd$
This is pretty trivial but isn't usually given as one of the standard inequality rules.
$\boldsymbol{Proof}$:
Given $a \ge c$ and $b \ge d$
Since $a \ge c$, we know that $ad \ge cd$
Since $b \ge d$, we know that $ab \ge ad$
So by transitivity we know that $ab \ge cd$
$\boldsymbol{Induction}$:
Now we can use induction to prove that given e and f are positive real numbers
(eqn 2)
$e \ge f \implies e^n \ge f^n$
(eqn 3)
Given $e \ge f$
Base case, n=1 is trivial:
Since $e \ge f$
$e^1 \ge f^1$
So true for n=1.
Assume true for $n=k$
(eqn 4)
$e^k \ge f^k$
Now we prove it's true for $n=k+1$
Using (3) and (4) and rule (1), we know that
$e^k(e) \ge f^k(f)$
$e^{k+1} \ge f^{k+1}$
That completes our proof of eqn 2.
$\boldsymbol{Substitution}$:
Now we can substitute $e = x^{1/n}$ and $f= y^{1/n}$ into eqn 2, where x and y are positive real numbers. Since x and y are positive, e and f are positive (we are taking the positive principal nth root).
$x^{1/n} \ge y^{1/n} \implies (x^{1/n})^n \ge (y^{1/n})^n$
$x^{1/n} \ge y^{1/n} \implies x \ge y$
Now we can take the contrapositive:
$x < y \implies x^{1/n} < y^{1/n}$
A: HINT:
$$x^{\frac{1}{n}}>y^{\frac{1}{n}}$$
What happens if we raise everything to the power of
$$\frac{n}{n+1}\;\;?$$
