Given $\Delta ABC$ is such that $a^3+b^3=c^3$, prove that $\Delta ABC$ is acute.
My erroneous working:
By the triangle inequality, $a+b>c \Rightarrow (a+b)(a^2-ab+b^2)>c(a^2-ab+b^2)$. We can do this since $a^2-ab+b^2>=ab>0$ by the $AM-GM$ Inequality.
Expanding yields $c^3>c(a^2-ab+b^2) \Rightarrow c^2>a^2-ab+b^2>a^2+b^2$, which is the condition for a triangle to be obtuse.
Can we generalise for $a^n+b^n=c^n$?
On
we distinguish two cases: we assume that the triangle is not accute, then we have in the first case the triangle hase a right angle, we assume we have $$a^2+b^2=c^2$$ or $$\sqrt{a^2+b^2}=c=\sqrt[3]{a^3+b^3}$$ powering by $6$ and rearranging we get $$2=3\frac{a}{b}+3\frac{b}{a}$$ Setting $$t=\frac{a}{b}$$ then we have the quadratic equation $$2t=3t^2+3$$ or $$t^2-\frac{2}{3}t+1=0$$ this equation hase no solution. Analogously we can solve the case that $$a^2+b^2<c^2$$
Using notations $\frac a c = x, \frac b c = y$ we have $x^3 + y^3=1$ where $0 \lt x, y \lt 1$. Because $x^2 \gt x^3, y^2 \gt y^3$ it follows $x^2 + y^2 \gt 1$. One can easily extend this proof to $a^n+b^n=c^n$