Let the \(P\left( n \right)\) be the statement:\({a^n} + {b^n} < {c^n}\).
Then by principle of mathematical induction,
For \(n = 3\):
\({a^3} + {b^3} < {c^3}\).
Thus, the result is true for \(n = 3\).
Hence, \(P\left( 3 \right)\) is true.
Since \(c\) is the length of the hypotenuse, it is the longest side of the triangle.
Therefore, it is written as:
\(\begin{aligned}{l}a < c,\,\,b < c\\ \Rightarrow \frac{a}{c} < 1,\,\,\frac{b}{c} < 1\end{aligned}\)
It will prove the further result by taking limit. Thus,
\(\begin{aligned}{c}\mathop {\lim }\limits_{n \to + \infty } \frac{{{a^n} + {b^n}}}{{{c^n}}} &= \mathop {\lim }\limits_{n \to + \infty } \frac{{{a^n}}}{{{c^n}}} + \mathop {\lim }\limits_{n \to + \infty } \frac{{{b^n}}}{{{c^n}}}\\ &= \mathop {\lim }\limits_{n \to + \infty } {\left( {\frac{a}{c}} \right)^n} + \mathop {\lim }\limits_{n \to + \infty } {\left( {\frac{b}{c}} \right)^n}\\ &= 0 + 0\\ &= 0\end{aligned}\)
Since \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{a^n} + {b^n}}}{{{c^n}}} = 0\) is possible only if \({a^n} + {b^n} < {c^n}\) for sufficiently large \(n\).
It knows that \({a^n} + {b^n} < {c^n}\) does not hold for \(n = 2\) because Pythagorean Theorem \({a^2} + {b^2} = {c^2}\) holds in equality for right triangles.
It notes that \({c^n}\) grows faster than \({a^n} + {b^n}\) while \({a^2} + {b^2} = {c^2}\).
This implies that \({a^n} + {b^n} < {c^n}\) for all positive integers \(n \ge 3\).
Hence, by the principle of mathematical induction, the result \(P\left( n \right)\) is true for all positive integers \(n \ge 3\).
