Question

1.Polynomial with positive ceofficients have only positive roots.

2. For any three elements a,b,c of ]0,1[ (1-a)(1-b)(1-c)>=1-(a+b+c)

Answer 1

1.

Let $P_k(x) = P(P(...(x))$ with k iterations of

we suppose (for contradiction) that P(x) has a non-positive root and let r be a positive root of P(x) so that $P(x) = (x - r)(x - s)f(x).$

Then $P_2(x) = (P(x) - r)(P(x) - s)g(x)$. Note that P(x) must have arbitrarily large negative values or arbit. large positive values to the left of s, thus

P(x) = r or P(x) = s for some x $\le$ s $\le$ 0

Now $P_3(x) = (P_2(x) - r)(P_2(x) - s)h(x)$ and repeating the same argument, we inductively prove that each $P_k(x)$, and hence $Q(x)$, must have both a positive root and a non-positive root, giving the desired contradiction.

2

For any three elements a, b, c of [0, 1] (1-a)(1-b)(1-c) $\geqslant$ 1-(a + b + c)

(1-a)(1-b)(1-c) $\geqslant$ 1-(a + b + c)

(1 - a - b + ab)(1 - c) $\geqslant$ 1 - (a + b + c)

1 - a - b + ab - c + ac + bc - abc $\geqslant$ 1 - (a + b + c);

ab + ac + bc - abc $\geqslant$ 0;

ab + ac + bc $\geqslant$ abc

we can prove this inequality for given numbers in [0, 1]:

for example, a = b = c = 1 give the correct inequality

3  $\geqslant$ 1

for example, a = b = c = 0 give the correct equality, it satisfies conditions of the inequality for minimum numbers in [0, 1]:

0 = 0

If 0<a<1, 0<b<1, 0<c<1, then the following inequality will be true

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 1$