Line under curve on the interval (proof)

Question

What is the easiest way how to prove that $3x+y=1$ is under $y=(1-x)^{3}$ without drawing graph? I mean by using some matematical technique. It is enough for interval $[-1,1]$. Thank you for advice.

Answer 1

$3x+y=1$ can be written as $y=1-3x$

We have to solve $$1-3x<(1-x)^3$$ Expand and rearrange $$1-3x<1-3x+3x^2-x^3$$ $$x^3-3x^2<0$$ $$x^2(x-3)<0$$ $$x<3$$ The line is under the curve for $x<3$

Answer 2

HINT

You have to prove that the graph of $y=(1-x)^3$ is over the graph of $y=1-3x$ over $x \in [-1,1]$. It suffices to prove that $$ \begin{split} (1-x)^3 &\ge 1-3x \\ 1 - 3x + 3x^2 - 3x^3 &\ge 1 - 3x \\ 3x^2 - 3x^3 &\ge 0 \end{split} $$ Can you finish?