$f(x)=x^3+ax^2+bx+c$ where $1\ge a\ge b\ge c\ge 0$. If $\lambda$ is any root of the polynomial, show that $|\lambda|\le 1$ $f(x)=x^3+ax^2+bx+c$ where $1\ge a\ge b\ge c\ge 0$. If $\lambda$ is any root of the polynomial, show that $|\lambda|\le 1$.  
My attempt:
As the polynomial is a cubic, it must have atleast one real root. So, I consider the $\mid\lambda\mid>1$ and $\lambda$ to be real and show that there arises a contradiction.  
Now, I differentiated the polynomial to see that $f'(x)>0$ from which I could conclude that if the roots are equal then I am done, but if roots are distinct then the other roots are imaginery. But, after several attempts, I could not show that $\mid\lambda\mid\le 1$. I considered using $\lambda\lambda'=\mid\lambda\mid^2$ where $\lambda'$ is the conjugate of the imaginery root, but that didn't seem to help. Please help.  
EDIT: Rouche's theorem is too advanced for me, please consider giving a more elementary solution. My level of knowledge should be near the undergraduate level.

Hint:
$$
\begin{align}
|(x-1)f(x)| &= \left|x^4 - \Bigl((1-a)x^3 + (a-b)x^2 + (b-c)x + c\Bigr)\right| \\
&\ge \left|x^4\right| - \left|(1-a)x^3 + (a-b)x^2 + (b-c)x + c\right| \\
&\ge |x|^4 - (1-a)|x|^3 - (a-b)|x|^2 - (b-c)|x| - c. \\
\end{align}
$$
Now suppose $|x| > 1$ and $f(x) = 0$ and obtain a contradiction.
Second hint: If $|x| > 1$ and $a < b$ then $|x|^a < |x|^b$.
