Induction for Inequalities II
Bernoulli's inequality is one of the most elegant results in analysis: $(1+x)^n \geq 1+nx$ for any $x > -1$. It underpins compound interest, exponential growth bounds, and calculus estimates. In this lesson you'll prove it by induction and tackle a classic series inequality — levelling up your manipulation skills along the way.
A savings account grows at 5% per year: $(1.05)^n$ times the original. Bernoulli's inequality says $(1.05)^n \geq 1 + 0.05n$. Test this for $n = 10$ using your calculator. Is the Bernoulli bound tight or loose? What does it tell you about how interest really compounds?
Bernoulli's inequality is a fundamental result used in analysis, finance and calculus. It provides a linear lower bound for a power function:
For any real $x > -1$ and any integer $n \geq 0$:
$(1+x)^n \geq 1 + nx$
The condition $x > -1$ ensures that $(1+x)$ is positive. This matters because in the inductive step we will multiply both sides of the hypothesis by $(1+x)$ — and multiplying by a positive number preserves the inequality direction.
At $n=0$: both sides equal 1 (equality). For $n \geq 1$ the power grows faster than the linear expression, so we get strict inequality when $x \neq 0$.
Key facts
- Bernoulli's inequality: $(1+x)^n \geq 1+nx$ for $x > -1$, $n \geq 0$
- The base case for Bernoulli is $n = 0$ (both sides equal 1)
- The comparison method: to show $X \leq Y$, prove $Y - X \geq 0$
Concepts
- Why multiplying both sides by $(1+x) > 0$ preserves the inequality
- Why the term $kx^2 \geq 0$ allows us to discard it to weaken the inequality
- The comparison method in series inequality proofs
Skills
- Write a complete proof of Bernoulli's inequality by induction
- Use the comparison method to prove series inequalities
- Justify every inequality direction change in a proof
Lessons 11 covered the basic “bridge the gap” technique. This lesson adds two further tools:
- Multiplying by a positive factor. In Bernoulli's proof, we multiply both sides of $(1+x)^k \geq 1+kx$ by $(1+x)$. Because $x > -1$ means $1+x > 0$, the inequality is preserved. This requires explicitly stating the condition.
- Discarding non-negative terms. After multiplying out, we get $(1+x)^{k+1} \geq 1 + (k+1)x + kx^2$. Since $kx^2 \geq 0$, discarding it gives $(1+x)^{k+1} \geq 1 + (k+1)x$. Must state $kx^2 \geq 0$; cannot just drop it silently.
Lessons 11 covered the basic “bridge the gap” technique. This lesson adds two further tools:
Pause — copy the two advanced inequality induction techniques (AM-GM application and multiplier comparison) with their trigger phrases into your book.
Quick check: In the proof of Bernoulli's inequality, after applying the inductive hypothesis you obtain $1+(k+1)x+kx^2$. What allows you to conclude $1+(k+1)x+kx^2 \geq 1+(k+1)x$?
Worked examples · 2 full proofs, reveal step by step
Prove by mathematical induction that $(1+x)^n \geq 1+nx$ for all $x > -1$ and all integers $n \geq 0$.
Prove by mathematical induction that $\displaystyle\sum_{r=1}^{n} \frac{1}{r^2} \leq 2 - \frac{1}{n}$ for all positive integers $n$.
Did you get this? True or false: in the proof of Bernoulli's inequality, the inductive step requires multiplying both sides of the hypothesis by $(1+x)$. This is only valid because $x > -1$ guarantees $1+x > 0$.
We just saw the basic "bridge the gap" inequality technique from Lesson 11. That raises a question: what two further tools — AM-GM style arguments and multiplier comparisons — handle more demanding inequalities that the basic technique cannot close? This card answers it → each tool has a specific trigger phrase and template proof structure.
- Bernoulli's inequality: $(1+x)^n \geq 1+nx$ for $x > -1$ and $n \geq 0$. Base case at $n = 0$ (equality).
- Inequality preservation: Multiplying by a positive number preserves the inequality direction. Always state why the multiplier is positive.
- Discarding positive terms: If $A = B + C$ and $C \geq 0$, then $A \geq B$. Must justify $C \geq 0$ explicitly.
- Comparison method: To show $X \leq Y$, prove $Y - X \geq 0$. Especially useful for series inequalities.
Key concept: Key concepts summary.
Pause — copy the complete decision guide for choosing among the three inequality induction techniques into your book.
Fill the gap: In the Bernoulli proof, after expanding $(1+kx)(1+x)$ you obtain $1 + (k+1)x + kx^2$. Since $kx^2 \geq $ , we can discard it to get $1+(k+1)x$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: if the base case of an inequality proof gives equality (e.g. LHS = RHS), the proof is invalid and you must find a different base case.
Activities · practice with the ideas
Verify the base case ($n=0$) of Bernoulli's inequality. State what happens when $x = 0$.
In the Bernoulli proof, write out the expansion of $(1+kx)(1+x)$ in full and identify the term that will be discarded.
Use Bernoulli's inequality to find a lower bound for $(1.1)^{20}$ without a calculator. Show your working.
Prove by induction that $\dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}} \geq \sqrt{n}$ for all integers $n \geq 1$. (Hint: show $\sqrt{k+1} - \sqrt{k} \leq \dfrac{1}{\sqrt{k+1}}$.)
State whether the following is valid in an induction proof: "Multiplying both sides by $(1-x)$ preserves the inequality." Justify your answer.
Odd one out: Three of these statements about Bernoulli's inequality are correct. Which one is NOT?
Earlier you computed $(1.05)^{10}$ and compared it to $1 + 0.05 \times 10 = 1.5$.
$(1.05)^{10} \approx 1.629$, which is greater than $1.5$. Bernoulli gives the correct inequality direction but not the exact value — it is a lower bound, not an exact formula. The real power of Bernoulli is that it works for all $x > -1$ and all $n \geq 0$, making it useful for proofs where you need a guarantee, not a precise calculation.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Apply Bernoulli's inequality to find a lower bound for $(1.2)^5$. Show your substitution clearly. (1 mark)
Q2. Prove by mathematical induction that $(1+x)^n \geq 1+nx$ for all $x > -1$ and all integers $n \geq 0$. (4 marks)
Q3. Prove by mathematical induction that $\dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}} \geq \sqrt{n}$ for all integers $n \geq 1$. (4 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. LHS $= (1+x)^0 = 1$, RHS $= 1 + 0 = 1$; $1 \geq 1$ ✓. When $x=0$: both sides are $1$ for all $n$.
2. $(1+kx)(1+x) = 1 + x + kx + kx^2 = 1 + (k+1)x + kx^2$. Discarded term: $kx^2$.
3. $x = 0.1$, $n = 20$: $(1.1)^{20} \geq 1 + 20 \times 0.1 = 3$.
4. Base: $n=1$: $1 \geq 1$ ✓. Hyp: sum $\geq \sqrt{k}$. Step: add $\frac{1}{\sqrt{k+1}}$. Need $\sqrt{k} + \frac{1}{\sqrt{k+1}} \geq \sqrt{k+1}$, i.e. $\frac{1}{\sqrt{k+1}} \geq \sqrt{k+1} - \sqrt{k} = \frac{1}{\sqrt{k+1}+\sqrt{k}} \leq \frac{1}{\sqrt{k+1}}$ ✓.
5. Only valid if $1-x > 0$, i.e. $x < 1$. Multiplying by a negative reverses the inequality, so must state the sign condition.
Q1 (1 mark): $x=0.2$, $n=5$: $(1.2)^5 \geq 1 + 5 \times 0.2 = 2$ [1].
Q2 (4 marks): Base $n=0$: $1 \geq 1$ ✓ [1]. Hyp: $(1+x)^k \geq 1+kx$ [1]. Step: since $x>-1$, $1+x>0$; multiply: $(1+x)^{k+1} = (1+x)^k(1+x) \geq (1+kx)(1+x) = 1+(k+1)x+kx^2$. Since $kx^2 \geq 0$: $\geq 1+(k+1)x$ [2].
Q3 (4 marks): Base $n=1$: $1 \geq 1$ ✓ [1]. Hyp: $\sum_{r=1}^k \frac{1}{\sqrt{r}} \geq \sqrt{k}$ [1]. Step: add $\frac{1}{\sqrt{k+1}}$; need $\sqrt{k} + \frac{1}{\sqrt{k+1}} \geq \sqrt{k+1}$. Aux: $\sqrt{k+1}-\sqrt{k} = \frac{1}{\sqrt{k+1}+\sqrt{k}} \leq \frac{1}{\sqrt{k+1}}$, so $\sqrt{k}+\frac{1}{\sqrt{k+1}} \geq \sqrt{k+1}$ ✓ [2].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering Bernoulli and series inequality questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.