I am doing problems that involve inequalities. My understanding is that you through a string of inequalities show that one is less than the other. Kind of like the transitive property. For example:
$2n+1 \lt 2^n$ for $n=3,4,...$
Assume this is true for P(k):
Base case $k = 3$
$LHS: 7 \space \space \space RHS: 8$
$LHS \space \space \lt \space \space RHS$
This holds true for $(k)$
Prove for P(k+1):
$2(n+1)+1 \lt 2^{n+1}$
my understanding is that if I can find something that is greater than $2(n+1)+1$ and obviously below $2^{n+1}$ then the inequality holds true
$2k+3 \lt 2^{k+1}$
$(2k + 1) + 2 \lt 2^k2$
$(2k+1) + 2 \lt 2^k + 2$ by our hypothesis on $k$
It is very obvious that $2^k + 2 \lt 2^k2$ and doesn't need explanation so its safe to assume
$2k+3 = (2k+1)+2 \lt 2^k + 2 \lt 2^k2$
Thus $(2k+3) \lt 2^k2$
This seems perfectly logical to me since we are dealing with integers. These inequalities would not hold for $\mathbb{R}$.
I am having a hard time find a string of inequalities that proves $n^2 \leq 2^n+1$
Assume this holds true for $k$. $P(k) = k^2 \leq 2^k+1$ for $n = 1,2...$
Base case $P(1):$ $LHS: 1 \space \space \space RHS: 3$
$LHS \space \space \leq \space \space RHS$
holds true for $P(k)$
$P(k+1)$:
$(k+1)^2 \leq 2^k + 1$
$k^2 + 2k + 1 \leq 2^k2+1$
$k^2 + 2k + 1 \leq 2^k +1 + 2k +1 \leq 2^k2+1$
$k^2 + 2k + 1 \leq 2^k + 2k + 2 \leq 22^k+1$ by our hypothesis on $k$
I do not know where to go from here