Prove that $x^{1/n} > x^{1/(n+1)}$ given that $x>1$.

Question

Prove that, for any $x > 1$:$$x^{1/n} > x^{1/(n+1)}$$ It looks like induction to me but since there's no equality, I don't know what to substitute in anywhere.

I can't see how this is an analysis question. what are you allowed to assume to prove this fact? You can show for $x>1$, $x^a>x^b$ on both side for $a>b$. then you divide $x^b$, then you say x = 1+c $(1+c)^{(a-b)}\geq 1+c(a-b)$ by Bernoulli. — Lost1, Sep 04 '13 at 23:01

score 7 · Accepted Answer · answered Sep 04 '13 at 22:58

7

Hint: Take the $n(n+1)$-th power of both sides.

Or, somewhat more simply, note that $x^{n+1}\gt x^n$. Raise each side of this inequality to the power $\frac{1}{n(n+1)}$.

The first way is more natural, the second a little slicker. They are equivalent.

answered Sep 04 '13 at 22:58

André Nicolas

507,029

score 5 · Answer 2 · answered Sep 04 '13 at 23:05

5

Taking logarithms will give you the equivalent inequality $\frac{\ln x}{n}>\frac{\ln x}{n+1}$, and this is true since $\ln x>0$.

answered Sep 04 '13 at 23:05

user84413

27,211

score 3 · Answer 3 · answered Sep 04 '13 at 23:25

3

Can we take for granted that $x\gt1$ implies $x^{1/N}\gt1$ for any $N$? If so, then

$${1\over n}-{1\over n+1}={1\over n(n+1)}$$

implies

$$x^{1/n}=x^{({1\over n}-{1\over n+1}+{1\over n+1})}=x^{1/n(n+1)}\cdot x^{1/ (n+1)}\gt 1\cdot x^{1/(n+1)}$$

answered Sep 04 '13 at 23:25

Barry Cipra

79,832

Prove that $x^{1/n} > x^{1/(n+1)}$ given that $x>1$.

3 Answers3