Prove $n+\left(-1\right)^n\ge \frac{n}{2}$ is true for $n\ge 2$ using induction

Question

I'm trying to prove $n+\left(-1\right)^n\ge \dfrac{n}{2}$ is true for all natural numbers $n \ge 2$ via induction. The base case is trivial as $$2+(-1)^2 \ge \frac{1}{2}(2)$$ $$3 \ge 1.$$ For the induction step, I'm looking at $$(n+1) + (-1)^{n+1} = n+1+(-1)(-1)^n$$ $$\ge \frac{1}{2}n +1-2\cdot(-1)^n$$ This is where I get stuck. Any help would be appreciated.

Answer 1

What Angina Seng hinted in comment is: $$n+(-1)^n\ge n-1\ge \frac n2 \iff n-\frac n2\ge 1 \iff n\ge 2.$$ To continue your line of thought, you must prove: $$n+1+(-1)^{n+1}\ge \frac{n+1}2 \iff \frac n2\ge (-1)^n-\frac 12 \iff \\ \frac n2\ge 1\ge (-1)^n-\frac12.$$

Answer 2

$$p(1):n=2 \to 2+(-1)^2\geq \frac 22\\p(n):n+(-1)^{n}\geq \frac{n}{2}\\p(n+1):n+1+(-1)^{n+1}\geq \frac{n+1}{2}$$ divide into two cases (n=even- odd) first case $n=even=2k ,k\geq 1$ $$n=even=2k \to p(n):2k+1\geq k\\p(n+1):2k+1+(-1)^{2k+1}\geq \frac{2k+1}{2}\\2k\geq k+\frac12 \checkmark (w.r.t. k\geq1)$$ second case $n=odd=2k+1,k\geq 1$ $$n=odd=2k+1 \to p(n):2k+1+(-1)^{2k+1}\geq \frac{2k+1}{2}\\p(n+1):2k+1+1+(-1)^{2k+2}\geq \frac{2k+2}{2}\\2k+3\geq k+1 \checkmark (w.r.t. k\geq1)$$