How prove this Polynomial $g(x)=\sum_{i=1}^{n}a^m_{i}x^i$have only real roots? Question 1:
let Polynomial $f(x)=\displaystyle\sum_{i=0}^{3}a_{i}x^i,$ have three real numbers roots,where $a_{i}>0,i=1,2,3$.

show that:
$$g(x)=\sum_{i=0}^{3}a^m_{i}x^i$$ have only real roots,where $m\in R,m\ge 1$

My try:
case (1):
suppose that $f(x)$  has  a zero of multiplicity 3, then we assume

$$f(x)=(x+p)^3=x^3+3x^2p+3xp^2+p^3$$
then
$$g(x)=x^3+(3p)^mx^2+(3p^2)^mx+(p^3)^{m}=(x+p^m)[x^2+(3^mp^m-p^m)x+p^{2m}]$$

then

$$h(x)=x^2+p^m(3^m-1)x+p^{2m}\Longrightarrow \Delta =(p^m(3^m-1))^2-4p^{2m}>0$$
so this case $g(x)$ have only three real roots.

for case (2):
let $f(x)=(x+p)^2(x+q)$,
I can't prove it,
and the case (3):
$$f(x)=(x+p)(x+q)(x+r)$$
and this case I can't prove it too.
I hope someone can help solve this nice problem ;
Thank you very much!
 A: It is really a tough question. A generalization of this problem follows from a theorem in the famous book by Polya and Szego(Book II, Chapter 5, Problem 155), which states:
If
$$a_0+a_1 x+\cdots+a_n x^n$$
and
$$b_0+b_1 x+\cdots+b_n x^n$$
have only real zeros, while all the zeros of the latter have the same sign, then
$$a_0 b_0+a_1 b_1 x+\cdots+a_n b_n x^n$$
has real zeros only.
The proof is complicated, so borrowing a copy from library is recommended.
A: As I explained in my comment, it is enough to do case (2).
Let $x,y \ge 0$. Then $(X+x)(X+x)(X+y) = X^3+(2x+y)X^2 + (x^2+2xy)X + x^2y$.
The discriminant of $X^3+bX^2+cX+d$ is, according to wikipedia, $\Delta(b,c,d) = b^2c^2-4c^3-4b^3d-27d^2+18bcd$.
The derivative of $m \mapsto \Delta(b^m,c^m,d^m)$ at $m=1$ is 
$$\delta(b,c,d) \\ 
= (2b^2c^2-12b^3d+18bcd)\log b + (2b^2c^2-12b^3d+18bcd)\log c + (18bcd-4b^3d-54d^2)\log d
$$
Plugging $b = 2x+y,c = x^2+2xy,d = x^2y$ we compute and obtain
$$ \delta(x,y) = 
 4x^6(\frac yx -1)^3\left(-(2+ \frac yx)\log \frac {(2x+y)^2}{x^2+2xy} + \frac yx \log \frac {(2x+y)^3} {x^2y}\right)$$
$\delta(x,y)/4x^6$ is actually a function of $t = \frac yx$ and has the same sign as $\delta(x,y)$ so it is enough to study the sign of the function $$g(t) = (t-1)\left(-(2+t)\log \frac{(2+t)^2}{1+2t} + t \log \frac {(2+t)^3}t\right) \\
 = (t-1)\left((t-4)\log(2+t) + (2+t)\log(1+2t) - t\log t\right)
$$
Letting $g(t) = (t-1)h(t)$, we compute $h'(t) = \log(9+(t- t^{-1})^2) + 2\frac{(t-1)^2}{2t^2+5t+2} \ge \log 9 > 0 $. Since $h(1) = -3\log3+3\log3-\log1 = 0$, we have $h(t)>0$ for $t>1$, $h(t)<0$ for $t<1$, and $h(1)=0$. 
Going back to $g$ we learn that $g(t)>0$ except when $t=1$, which means that $\delta(x,y) > 0$ except when $x=y$, which is what we wanted.
A: For the first question, this is really a question about the derivative. $f$ has three real zeros if and only if its derivative has two zeros, and $f$ is positive at the smaller critical point, and negative at the bigger critical point [by the way, I am dividing through by $a_3$ to make the polynomial monic -- this does not change the question). This can be checked for $g$ by tedious computation, which I leave to you...
