Proving Inequalities using Induction

Induction hypothesis is not $2^k\geq 2k$ but $k^2\geq 2k$. Then, for $P(k+1)$, we have to prove $(k+1)^2\geq 2(k+1)$.

Proof:

$(k+1)^2=k^2+2k+1$ but $k^2 \geq 2k$ (by IH) $\implies k^2+2k+1\geq (2k+2k+1=4k+1)\geq 2k+2$ as $k\geq 1\implies (k+1)^2\geq 2(k+1)$. Hence ,$P(k+1)$ is true whenever $P(k)$ is true. Since $P(1)$ is true $\implies P(n)$ is true $\forall n\in \Bbb Z^+$.

We have to prove $2^n \geq 2n$ for $n>1$
Basis:
$n=2$ which satisfies the above relation.

Induction hypothesis:
Here we assume that the relation is true for some $k$ i.e. $P(k)\colon 2^k \geq2k$.

Now we have to prove that the relation also holds for $k+1$ by using the induction hypothesis. This means that we have to prove $$P(k+1)\colon 2^{k+1} \geq 2(k+1)$$ So the general strategy is to reduce the expressions in $P(k+1)$ to terms of $P(k)$. So,
$$2^{k+1}=2^k\cdot 2=2^k+2^k \geq2^k+2 \quad \quad \{\text{Since }n>1\}$$ Now we will use induction hypothesis that $2^k\ge 2k$ which gives us $$ 2^{k+1}\geq 2^k+2 \geq 2k+2 =2(k+1)$$ which was required. Hence we have proved $P(k+1).$