$(1+x)^n \ge \frac{n(n-1)}{2}x^2$ is there something wrong in my proof?

Question

This problem comes from my textbook, and I think it has mistake with equality sign, but if it doesn't, please, give me some hints. Here is the problem. It asks me to show inequality by using binomial theorem

And here is my steps

as cannot be ZERO i think there shouldn't be equality sign. maybe x should have been choosen as -1 and n=1 for example? Thanks and sorry for bad english.

Answer 1

Your proof is correct, just observe that if $a>0$ is true then $a\geq0$ is true, in the latter you have $a=0$ $\color{red}{or}$ $a>0$.

Answer 2

It does seem that the problem was posed in a less ambitious form than it might have been.

The only problem I can find in your proof (and it is very small and easily fixed) is that you appear to assume implicitly that $N \geq 2.$ If $N < 2$ then you have to explain what you mean by writing $\binom n2$ in that case.

One thing you can do is prove three separate cases: $N=0,$ $N=1,$ and $N\geq 2.$ The first two cases are trivial, and then you are left with the third case which says explicitly that $N\geq 2$ and you can obviously write $\binom n2$.

Since you can prove that a strict inequality holds, it follows that the non-strict inequality holds as well.

Answer 3

Check $x=−2$ and $n$ being positive odd in the statement you are trying to prove.