Prove $(x^n+1)^{\frac 1n}≤ x+1$. x>0 What is wrong with my proof?

This is what I did so far: I shall prove this by induction on $n$.

Let $n=1$, then $x+1≤x+1$ which is correct.

Let the inequality hold for $n$.

Then, $(x^n+1)^{\frac 1n}≤x+1$, e.g, $x^n+1≤(x+1)^n$. I'll prove that it holds for $n+1$:

$(x^{n+1} + 1)^{\frac{1}{n+1}}≤ x + 1 => x^{n+1} + 1≤(x + 1)^{n+1} => x*x^n + 1≤(x + 1)^n*(x+1)$

My question is: can I replace $(x + 1)^n$ by $(x^n+1$) which follows from my assumption?

The problem is that I put something smaller in the side that is supposed to be bigger which might be problematic. Of course it is not hard to make sure by hand that making the replacement will not change the inequality, but still, is it proper process?

• Use LaTex man (or woman). – barak manos Jan 13 '15 at 12:02
• How (based on what) do you conclude $(x^{n+1} + 1)^{\frac 1n}≤ x + 1$? – barak manos Jan 13 '15 at 12:09
• In order to prove by induction, you have to start with the $n+1$ proposition and show, using the fact that the property holds for $n$, that it holds also for $n+1$. You did not do this, you assumed that it did and tried to get back on your feet afterwards. – Martigan Jan 13 '15 at 12:09
• I don't conclude... It's just an inequality waiting to either be true or false...I could as well adding "?" all along – Meitar Jan 13 '15 at 12:10
• Hint: Multiply both sides of the statement $P(n)$ with $(x+1)$ and bound one side to show it implies $P(n+1)$ holds... – Macavity Jan 13 '15 at 12:17

There is a logical misstep waiting to happen here. If you start from the proposition you want to prove and work your way to a statement you know is true, you have not proven your proposition. The logical inference you attempt to make is essentially the following: $Q$ is true and $P\Rightarrow Q$, therefore $P$ is true, which is an invalid inference! The correct inference reads: $P$ is true and $P\Rightarrow Q$, therefore $Q$.
You therefore have to start from the statement you know is true: $x^n+1\leq (x+1)^n$, and try to derive $x^{n+1}+1\leq (x+1)^{n+1}$. To do this, you might try to first multiply both sides by $(x+1)$, whence you obtain $x^{n+1}+x^{n}+x+1\leq (x+1)^{n+1}$. I'll let you take it from here.