If $a^2 + b^2 = c^2$, then $a^3 + b^3 < c^3$, for $a$, $b$, $c$ the sides of a triangle

Question

If $a$, $b$, $c$ are the sides of a triangle where $a^2 + b^2 = c^2$, prove that $a^3 + b^3 < c^3$.

I've tried triangle inequality, but I am stuck.

Answer 1

HINT:

As $\dfrac{a^2}{c^2}=1-\dfrac{b^2}{c^2}\le1\implies\dfrac ac\le1$

$$\implies\dfrac{a^3}{c^3}\le\dfrac{a^2}{c^2}$$

So, $$\dfrac{a^3}{c^3}+\dfrac{b^3}{c^3}\le\dfrac{a^2}{c^2}+\dfrac{b^2}{c^2}$$

but the equality occurs iff $a=c, b=c$

Answer 2

You have $c^2 = a^2 + b^2$. Now multiply by $c$ both sides and get: $c^3 = ca^2 + cb^2$. However you know that $c = \sqrt{a^2 + b^2}$, so in particular you know that $c > a$ and $c > b$. hence you know that $ca^2 > a^3$ and $cb^2 >b^3$. Therefore $c^3 > a^3 + b^3$.

Answer 3

If $c^2=a^2+b^2$, then: $$ \color{red}{c^6} = a^6+3a^4 b^2+3a^2 b^4 + b^6 \color{red}{>} a^6 + 2a^3 b^3 +b^6 = \color{red}{(a^3+b^3)^2} $$ since the quadratic form $3a^2-2ab+3b^2$ has a negative discriminant.