Verification for this proof Sorry guys about the verification questions but it's near the end of the semester and I am very sheepish about making mistakes especially because real analysis is a very important course it's only the underlying theory of calculus.
This problem may seem rudimentary but we were given a sample test by our professor and it is worth quite a bit of marks I need to make sure I'm not wrong.
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Here's the problem:
Prove: Prove $(1+x)^n ≥ 1+nx$. Where $x>-1$ and $n≥2$.
My Proof:
First we'll start off with the fact,
$(1+x)^2≥2x+1$ when $x>-1$ which is equivalent to $x^2>0$ which is obviously true for a real $x$.
So, by induction, we must prove if,
$(1+x)^n≥1+nx$
Than,
$(1+x)^{n+1}≥1+(n+1)x$.
Here's my proof:
$(x+1)^n(x+1)≥(1+x)+nx$
Therefore,
$(x+1)^n≥1+\frac{nx}{x+1}$.
If, $(x+1)^n≥1+nx$ then surely,
$(x+1)^n ≥ 1+\frac{nx}{x+1}$
Q.E.D
Any help or other method will be appreciated.

The reasoning you are using for your proof can be distilled to the following:

I want to prove that if a triangle has all angles equal, then its angles are all $60^\circ$. Here is my proof. Its angles are all $60^\circ$. Therefore its angles are surely all equal.

I hope it is clear why this reasoning is invalid.
One more thing. "Q.E.D." is quite useless.
