Exponential Comparison of Three Four and Five

Problem Statement

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

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.

Solutions

The statement is true.

Proof by Example:

Try $n = 3$:

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):

Divide the inequality by $5^n$:

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)
• 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: $3^2 + 4^2 = 5^2$).

Takeaways

• Reconstruct the full proof from the hint and the solution outline, and justify every transformation explicitly.
• Check edge cases and verify where each assumption is used in the argument.

Further Readings

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

• HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
• HSC Last Resorts: https://vumaths.com/booklets/hsc-last-resorts/
• HSC Trigonometry: https://vumaths.com/booklets/hsc-trigonometry/

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