SImple proof based on Inequalities

Question

In a right-angled $\triangle ABC$, which is right-angled at $C$, prove that $a^n + b^n < c^n$ for all $n > 2$.

I am able to prove this for powers of 2 as $a^2 + b^2 = c^2$

$$(a^2 + b^2)^n > (a^2)^n + (b^2)^n$$ But how to prove that it's true for all $n$ int.s

Try to divide by $c^2$. Then use trigonometric functions. – Andrei Mar 01 '22 at 16:59 — Andrei, Mar 01 '22 at 16:59

score 5 · Accepted Answer · answered Mar 01 '22 at 17:07

5

As $0<\frac{a}{c}<1$ and $0<\frac{b}{c}<1$, one has for $n>2$:

$$\left(\frac{a}{c}\right)^n+\left(\frac{b}{c}\right)^n<\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1$$

answered Mar 01 '22 at 17:07

Momo

1 Answers1