Induction is just as powerful for inequalities as for equalities, but the inductive step needs a slightly different reflex. With an equality you transform one side into the other; with an inequality you build a chain of $\le$ or $<$ steps from the $(k+1)$-expression down to (or up to) the target, using the hypothesis at one link.
The general shape of an inequality step. Suppose $P(n): A(n) \ge B(n)$. To prove $P(k+1)$ you typically start from $A(k+1)$, relate it to $A(k)$, apply the hypothesis $A(k) \ge B(k)$, and then show the resulting quantity is still at least $B(k+1)$:
$$A(k+1) \;\ge\; \big(\text{expression in } A(k)\big) \;\overset{\text{hyp}}{\ge}\; \big(\text{expression in } B(k)\big) \;\ge\; B(k+1).$$
Every link must point the same way. The most common slip is reversing an inequality when multiplying or when dropping a positive term, so state explicitly why each step is valid (e.g. "adding the positive quantity $2^k$" or "since $k \ge 1$").
Watch the starting value. Many inequalities are false for small $n$ and only become true past a threshold. The base case must be taken at the first $n$ for which the claim is asserted:
Bernoulli's inequality. A landmark application: for every real $x > -1$ and every $n \in \mathbb{N}$,
$$(1 + x)^n \ge 1 + nx.$$
Its inductive step is a clean model of the chain idea: $(1+x)^{k+1} = (1+x)(1+x)^k \ge (1+x)(1+kx)$ — valid because $1 + x > 0$ lets us multiply the hypothesis through — and expanding gives $1 + (k+1)x + kx^2 \ge 1 + (k+1)x$, since $kx^2 \ge 0$. The discarded term $kx^2$ is exactly what makes the inequality (not equality) hold.
Other applications. Induction underlies many results you meet later: defining a sequence recursively and proving a closed form for it, proving the number of subsets of an $n$-set is $2^n$, the number of diagonals of an $n$-gon is $\dfrac{n(n-3)}{2}$, and a host of inequalities used in calculus and analysis. The same skeleton applies; only the algebra of the step changes.
Deeper Insight — inequalities reward "spend less than you have". The art of an induction inequality is choosing how much of the hypothesis to use. You frequently prove something weaker than the tightest bound at each step (dropping a non-negative term such as $kx^2$ or $3k$) precisely because the weaker statement is still strong enough to reach $B(k+1)$ and is far easier to manipulate. Beginners often try to keep every term and get stuck; experts deliberately throw away what they do not need. Recognising that an inequality gives you slack — room to discard positive quantities — is the single insight that makes these proofs feel routine rather than mysterious.
Prove by induction that $2^n > n$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: $2 > 1$. True.
- Hypothesis: $2^k > k$ for some fixed $k \ge 1$.
- Step: $2^{k+1}=2\cdot2^k > 2k$ (multiplying the hypothesis by $2$).
- Since $k \ge 1$, we have $2k = k + k \ge k + 1$.
- Hence $2^{k+1} > k + 1$.
Answer: By PMI, $2^n > n$ for all $n\in\mathbb{N}$.
Prove by induction that $2^n \ge n + 1$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: $2 \ge 2$. True.
- Hypothesis: $2^k \ge k + 1$.
- Step: $2^{k+1}=2\cdot2^k \ge 2(k+1)=2k+2$.
- Since $k \ge 1$, $2k + 2 \ge k + 2 = (k+1)+1$.
Answer: By PMI, $2^n \ge n + 1$ for all $n\in\mathbb{N}$.
Prove by induction that $n! > 2^{\,n-1}$ for all $n \ge 3$.
Solution- Base: $n=3$: $3! = 6 > 2^{2} = 4$. True.
- Hypothesis: $k! > 2^{\,k-1}$ for some fixed $k \ge 3$.
- Step: $(k+1)! = (k+1)\cdot k! > (k+1)\cdot 2^{\,k-1}$.
- Since $k \ge 3$, $k + 1 \ge 2$, so $(k+1)\cdot 2^{\,k-1} \ge 2\cdot 2^{\,k-1}=2^{k}$.
- Thus $(k+1)! > 2^{k}=2^{(k+1)-1}$.
Answer: By PMI, $n! > 2^{\,n-1}$ for all $n \ge 3$.
Prove Bernoulli's inequality $(1 + x)^n \ge 1 + nx$ for all $n \in \mathbb{N}$ and $x > -1$.
Solution- Base: $n=1$: $(1+x)^1 = 1 + x \ge 1 + 1\cdot x$. True (equality).
- Hypothesis: $(1+x)^k \ge 1 + kx$.
- Step: since $1 + x > 0$, multiply both sides by $(1+x)$: $(1+x)^{k+1} \ge (1+x)(1+kx)$.
- Expand: $(1+x)(1+kx)=1+(k+1)x+kx^2 \ge 1+(k+1)x$, because $kx^2 \ge 0$.
Answer: By PMI, $(1+x)^n \ge 1 + nx$ for all $n\in\mathbb{N}$, $x > -1$.
Prove by induction that $2^n > n^2$ for all $n \ge 5$.
Solution- Base: $n=5$: $2^5 = 32 > 25 = 5^2$. True.
- Hypothesis: $2^k > k^2$ for some fixed $k \ge 5$.
- Step: $2^{k+1}=2\cdot2^k > 2k^2$.
- It suffices to show $2k^2 \ge (k+1)^2$, i.e. $k^2 - 2k - 1 \ge 0$, which holds for $k \ge 5$ (since $k^2-2k-1=(k-1)^2-2 \ge 14$).
- Hence $2^{k+1} > (k+1)^2$.
Answer: By PMI, $2^n > n^2$ for all $n \ge 5$.
Prove by induction that $n < 2^n$ for all $n \in \mathbb{N}$, and explain why the inequality is strict.
Solution- Base: $n=1$: $1 < 2$. True.
- Hypothesis: $k < 2^k$.
- Step: $k + 1 \le 2k < 2\cdot2^k = 2^{k+1}$ (first inequality uses $k \ge 1$, second uses the hypothesis).
- The hypothesis is strict and is preserved through the chain, so the result is strict.
Answer: By PMI, $n < 2^n$ for all $n\in\mathbb{N}$ (strict throughout).
Prove by induction that $1 + 2n \le 3^n$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: $1 + 2 = 3 \le 3^1 = 3$. True.
- Hypothesis: $1 + 2k \le 3^k$.
- Step: $3^{k+1}=3\cdot3^k \ge 3(1+2k)=3+6k$.
- We need $3 + 6k \ge 1 + 2(k+1)=2k+3$, i.e. $4k \ge 0$, true for $k \ge 1$.
Answer: By PMI, $1 + 2n \le 3^n$ for all $n\in\mathbb{N}$.
Prove by induction that the sum of the first $n$ positive integers satisfies $1 + 2 + \cdots + n \le n^2$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: $1 \le 1$. True.
- Hypothesis: $1 + 2 + \cdots + k \le k^2$.
- Step: $1 + 2 + \cdots + k + (k+1) \le k^2 + (k+1)$.
- Now $k^2 + k + 1 \le k^2 + 2k + 1 = (k+1)^2$, since $k + 1 \le 2k + 1$ for $k \ge 0$.
Answer: By PMI, $1 + 2 + \cdots + n \le n^2$ for all $n\in\mathbb{N}$.
Why must the base case for $n! > 2^{\,n-1}$ be taken at $n = 3$ rather than $n = 1$?
Solution- At $n = 1$: $1! = 1$ and $2^{0}=1$, so $1! > 2^{0}$ is false (they are equal).
- At $n = 2$: $2! = 2$ and $2^{1}=2$, again equal, so the strict inequality fails.
- At $n = 3$: $3! = 6 > 4 = 2^{2}$, the first value where it becomes true.
Answer: The claim only holds for $n \ge 3$, so the base case must be $P(3)$.
In proving $(1+x)^n \ge 1 + nx$, which term is discarded in the step, and why is that legitimate?
Solution- Expanding $(1+x)(1+kx)$ gives $1 + (k+1)x + kx^2$.
- The term $kx^2$ is dropped to reach $1 + (k+1)x$.
- Dropping it only weakens the right side, and $kx^2 \ge 0$ for all real $x$, so the inequality is preserved.
Answer: The non-negative term $kx^2$ is discarded; removing a non-negative quantity keeps the $\ge$ valid.