Prove $1+a0,b>0,c>0$, $aSuppose:
$a>0,b>0,c>0$, $a<bc$ , $1+ a^{3} = b^{3} + c^{3} $.
Prove:
$$ 1+a<b+c $$
This inequality at the University of Toronto plan.

First, note that $$1+a^3=b^3+c^3<1+(bc)^3$$ Which implies that $$(b^3-1)(c^3-1)>0 \Leftrightarrow (b-1)(c-1)>0$$ As we have $b^2+b+1, c^2+c+1>0$. Multiplying $3(b+c)$ on each sides, we have $$3bc(b+c)-3(b+c)^2+3b+3c>0$$ And addding $b^3+c^3-a^3-1$ on each side we have $$b^3+c^3+3bc(b+c)-3(b+c)^2+3(b+c)-1-a^3>0 \Leftrightarrow (b+c-1)^3-a^3>0$$
And since $x^3$ is a strictly increasing function we have $$b+c-1>a \Leftrightarrow b+c>a+1$$ As desired.
