$f(x)=x^3+ax^2+bx+c$ 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.
 A: 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 $|x|^a < |x|^b$.
A: 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. 
A: The standard way of solving this problem is using Rouche's theorem. Following 
are some alternatives.
Method 1 Companion matrix + Gershgorin circle theorem
Alternatively, one can look at the companion matrix $M$ associated with the polynomial
$$(x-1)f(x) = x^4 + (a-1)x^3 + (b-a)x^2 + (c-b)x-c$$
We have
$$M = \begin{bmatrix}
0 & 0 & 0 & c\\
1 & 0 & 0 & b-c\\
0 & 1 & 0 & a-b\\
0 & 0 & 1 & 1-a
\end{bmatrix}
\quad\text{ and }\quad
\det(x I_4 - M) = (x-1)f(x)
$$
If one apply Gershgorin circle theorem to $M$ along the columns, we immediately find 
the roots of $(x-1)f(x)$ lies on the union of two circle, one centered at $0$ with
radius $1$ and another one centered at $1-a$ with radius $|c|+|b-c|+|a-b| = a$.
The last circle is a subset of the first circle. This implies all roots of $(x-1)f(x)$
and hence those of $f(x)$ fall inside the closed unit circle.
Method 2 Descartes rules of signs + Vieta's formulas
By Descartes' rule of signs,
it is easy to check $f(x)$ has either one or three real roots in $[-1,0]$ and there are
no roots on other part of the real axis. 
If there are three real roots, we are done. Otherwise, two of the roots are complex. Let the three roots be $z = -\lambda + \mu i, \bar{z} = -\lambda - \mu i$ and $-\nu$ where $\lambda, \mu, \nu \in \mathbb{R}$. 
Since $-\nu$ is the only real root, we know $\nu \in [0,1]$. By 
Vieta's formulas, we have
$$
\begin{cases}
-a &= z + \bar{z} - \nu\\
 b &= (z + \bar{z})(-\nu) + z\bar{z}\\
-c &= z\bar{z}(-\nu)
\end{cases}
\quad\iff\quad
\begin{cases}
a &= 2\lambda + \nu\\
b &= 2\lambda\nu + \lambda^2 + \mu^2\\
c &= (\lambda^2 + \mu^2)\nu
\end{cases}$$
Combine the $1^{st}$ and $2^{nd}$ formulas, we have
$$\lambda^2 + \mu^2 = b + \nu(\nu - a) \le b + 1 - a \le 1 \quad\text{ because }\quad \nu \in [0,1]$$
This means the modulus of the two complex roots $z$ and $\bar{z}$ are also smaller than one.
A: Given that all the coefficients are non-negative and the lead coefficient is $1$, no real root can be positive. Suppose that $-\lambda$ is a real root and $\lambda\gt1$, then by the conditions given
$$
0=\lambda^3-a\lambda^2+b\lambda-c=\overbrace{(\lambda-a)}^{\gt0}\lambda^2+\overbrace{(b\lambda-c)}^{\ge0}\gt0
$$
Thus, $-1\le-\lambda\le0$.
That is, any real root must be in $[-1,0]$.

If we have only one real root, $\lambda\in[-1,0]$, then the two conjugate roots satisfy
$$
\frac{x^3+ax^2+bx+c}{x-\lambda}=x^2+(a+\lambda)x+(b+a\lambda+\lambda^2)=0
$$
Thus, the product of the conjugate roots, which is the square of their absolute values, is $b+a\lambda+\lambda^2$, which, because it is a convex function of $\lambda\in[-1,0]$, is less than its maximum at the endpoints of $[-1,0]$. That is, $b+a\lambda+\lambda^2\le\max(b,1-(a-b))\le1$.
That is, the absolute value of any complex root must be $\le1$.
A: since k(let consider)is a root of that equation we have


*

*$ k^3 = -ak^2-bk-c $

*this implies that
                 $ k^4=-ak^3-bk^2-ck $=$ (1-a)k^3+(a-b)k^2+(b-c)k+c $   [ where we again used $ k^3+ak^2+bk+c=0 $ ]

*suppose |k|$\geq$1.then we obtain $ |k|^4\leq (1-a)|k|^3+(a-b)|k|^2+(b-c)|k|+c\leq (1-a)|k|^3+(a-b)|k|^3+(b-c)|k|^3+c|k|^3 =|k|^3 $

*this shows that |k|$\leq$1.so the only possibility in this case is |k|=1.so we conclude that |k|$\leq$1 is always true.

A: I would recommend using Rouché's theorem. By splitting you function correctly you should be able to prove that all solutions $\lambda$ (real or complex) are such that $|\lambda| \le 1$.
You could split the function as $f(z) = z^3 + az^2$ and $g(z) = bz + c$ and compare their modulus on $|z| = 1$, $z = x+iy$
$$|f(z)| = |z^3 + az^2| = |z+a| =(x+a)^2 + y^2$$
$$|g(z)| = |bz + c| = (bx+c)^2 + y^2$$
Using the conditions on $a,b$ and $c$ you can show that $|f(z)| \ge |g(z)|\, \forall |z| = 1$ (try it).
Since $f(z) = 0$ has three zeros within $|z| = 1$, so does $f(z) + g(z) = z^3 + az^2 + bz + c$
That said there are other ways to prove it using Rouché's Theorem.
