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.

This proof goes by showing the assumption $1+a \ge b+c$ leads to a contradiction. Since the variables are positive the assumption is equivalent on squaring the sides to
$$1+2a+a^2 \ge b^2+2bc+c^2.\tag{1}$$
Now from $a<bc$ follows $-3a>-3bc,$ which on adding to (1) gives the strict inequality
$$1-a+a^2>b^2-bc+c^2.\tag{2}$$
Together with the assumption $1+a \ge b+c,$ since one of the inequalities is strict, we get to
$$(1+a)(1-a+a^2)>(b+c)(b^2-bc+c^2),$$
that is, to $1+a^3>b^3+c^3,$ against the assumption $1+a^3=b^3+c^3$ of the problem.
