$f(x)=x^3+ax^2+bx+c$ 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.

What you are looking for is known as the Enestrom-Kakeya Theorem. 
Thm. Suppose $g(x):=a_0+a_1x+\cdots+a_{n-1}x^{n-1}+a_nx^n$ is a polynomial with real positive coefficients of degree $n$. Then every root of $g$, $z_0\in\mathbb{C}$, has the following bounds, $$\min_{1\le k\le n} \left\{\frac{a_{k-1}}{a_k}\right\}=\delta \le |z_0| \le \gamma= \max_{1\le k\le n} \left\{\frac{a_{k-1}}{a_k}\right\}.$$
Proof. We need to show that the largest zero (in modulous) of $(x-\gamma)g(x)$ is $\gamma$. We also need to show that the largest zero (in modulus) of $(x-\frac{1}{\delta})x^n g(\frac{1}{x})$ is $\frac{1}{\delta}$. However, the second part is unnecessary for your cubic case. So, I will show just the first statement. 
Consider, $(x-\gamma)g(x)=a_n(x^n-b_1x^{n-1}-\cdots-b_n)$, where each $b_k$ will be positive, since $\gamma$ is the max of the ratios (literally multiply out the product). Define, 
$$f(x)=x^n-b_1x^{n-1}-\cdots-b_n.$$ 
Note that $\gamma$ is a zero of $f$. We wish to show that the largest zero of $f$ is $\gamma$. Consider the following function, 
$$F(x):=-\frac{f(x)}{x^n}.$$ 
Since $F$ is simply the sum of rational functions with a pole at only zero, we deduce that $F(x)$ is a strictly decreasing function, but since $\lim_{x\to 0^+} F(x)=\infty$ and $\lim_{x\to\infty} F(x)=-1$, we conclude that $F(x)$ has only one real zero (in fact of multiplicity one) in the interval $(0,\infty)$. But we already know the one real zero of $F$, it is $\gamma$. Thus, $F(x)$ is positive on $(0,\gamma)$ and negative on $(\gamma,\infty)$. 
Now suppose that $f(z)=0$ for some $z\in\mathbb{C}$, $z\not=0$. If $z$ happens to be zero, then there is nothing to prove since obviously, $|0|\le\gamma$. We now have, 
$$0=z^n-b_1z^{n-1}-\cdots-b_n,$$ thus $$z^n=b_1z^{n-1}+\cdots+b_n,$$ thus by triangle inequality, $$|z|^n\le b_1 |z|^{n-1}+\cdots+b_n,$$ thus dividing and minus one, $$0\le -1+\frac{b_1}{|z|}+\cdots+\frac{b_n}{|z|^n}=F(|z|).$$ Hence $F(|z|)\ge 0$, but we showed earlier that $F$ is negative on $(\gamma,\infty)$. Thus, $$0\le |z|\le \gamma.$$ That is, $\gamma$ is indeed the largest zero of $f$. 
Since you want to show that 1 is a largest bound for your cubic polynomial, try instead to show that the largest zero of $(x-1)p(x)$ is one itself. 
