Prove $1+a0,b>0,c>0$, $a

Question

Suppose:

$a>0,b>0,c>0$, $a<bc$ , $1+ a^{3} = b^{3} + c^{3} $.

Prove:

$$ 1+a<b+c $$

This inequality at the University of Toronto plan.

Answer 1

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.

Answer 2

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.