This is an application of holder's inequality. Since $(1+x^{n+1})^{\frac{1}{n}}(1+x)^{\frac{1}{n}}...(1+x)^{\frac{1}{n}} \geq 1+x^2$ and $ 1+x^2 \geq 2x$.

It's very easy.We have. 1+x^{n+1}\geq\ 2x^{\frac{n+1}{2}}; (1+x)^{n-1}\geq\ 2^{n-1}.x{\frac{n-1}{2};

Indeed, it is only second problem in the first day. It should be quite easy

Another thought is induction. For n=0 we have x+1=>x+1. We only have to prove that x^(n+3) - x^(n+2) - x +1 =>0 or that (x^(n+2) - 1) * (x-1)=>0 which is obvious!!! (the equality holds iff x=1) JJ-THE-KING

AM-GM

What about Bernoulli?