Exponential Comparison of Three Four and Five

In this problem we investigate an exponential inequality that blends algebra with calculus. It tests your ability to prove or disprove existence statements—a core skill in HSC Extension 1 and 2—and introduces the powerful technique of dividing by the largest term to transform a comparison. By working through the solution, you’ll see how a single numerical example can settle an existence question, while deeper analysis reveals the full picture and shows that the inequality holds for all real numbers beyond a critical point.

Problem Statement

Prove or disprove: There exists a real number $n$ such that $3^n + 4^n < 5^n$.

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Hints

The statement is true. Try a few values: $n = 2$ gives $9 + 16 = 25 = 5^2$ (equality, not $<$).

Try $n = 3$: Is $3^3 + 4^3 < 5^3$? Calculate $27 + 64$ vs $125$.

Alternatively, divide by $5^n$ to get $(3/5)^n + (4/5)^n < 1$ and analyze the function behavior.

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Solutions

The statement is true.

Proof by Example:

To prove that such a real number exists, we only need to find one value of $n$ that works. From the hint, we know $n=2$ gives equality, so we can test the next convenient integer. Let's try $n=3$:

$$\begin{aligned} 3^3 + 4^3 &= 27 + 64 = 91 \\ 5^3 &= 125 \end{aligned}$$

Since $91 < 125$, the inequality $3^3 + 4^3 < 5^3$ holds.

Therefore, $n = 3$ is a real number satisfying the inequality. $\blacksquare$

Alternative Proof (Analysis):

We now want to understand for exactly which $n$ the inequality is true. A clever first step is to divide through by the dominant term $5^n$, which transforms the inequality into a sum of fractions that are each less than 1.

Divide the inequality by $5^n$:

$$\left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n < 1$$

Let $f(n) = \left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n$.

• At $n = 2$: $f(2) = \frac{9}{25} + \frac{16}{25} = 1$ (equality)
• To see how $f(n)$ behaves beyond $n=2$, we examine its derivative:

The derivative: $f'(n) = \left(\frac{3}{5}\right)^n \ln\left(\frac{3}{5}\right) + \left(\frac{4}{5}\right)^n \ln\left(\frac{4}{5}\right)$

Since $\frac{3}{5} < 1$ and $\frac{4}{5} < 1$, both logarithms are negative, so $f'(n) < 0$.

• Therefore, $f(n)$ is strictly decreasing.

Since $f(2) = 1$ and $f$ is strictly decreasing, for any $n > 2$, we have $f(n) < 1$.

Thus, the inequality holds for all $n > 2$. In particular, it holds for $n = 3$ (and infinitely many other values). $\blacksquare$

Note: This shows the power of the "divide by largest term" technique in inequality problems. The critical point is $n = 2$ where equality holds (Pythagorean triple: $3^2 + 4^2 = 5^2$).

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Takeaways

• For an existence proof, a single concrete counterexample (like $n=3$) is enough—you don’t need to find all solutions.
• The “divide by the largest term” trick transforms an exponential comparison into a sum of fractions less than 1, making monotonic behaviour easier to analyze.
• Recognising when a function is strictly decreasing (via the derivative) reveals the full range of solutions: here, all $n>2$, with equality at the critical $n=2$—the familiar $3$-$4$-$5$ Pythagorean triple.

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Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Last Resorts, HSC Trigonometry