Solving equations $a+b+c=5,a^2+b^2+c^2=11,a^3+b^3+c^3=27$ 
Let $a, b, c $ be real numbers such that $a<b<c$ and satisfying
  $$a+b+c=5;$$
  $$a^2+b^2+c^2=11;$$
  $$a^3+b^3+c^3=27.$$
  Prove that $0<a<1<b<2<c<3$.

My understanding :
$(a+b+c)^2 - (a^2+b^2+c^2) = 14$
so, $ab+bc+ca = 7$
$a^2+b^2+c^2-ab-bc-ca = 4$
$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=20$
so, $abc = -\frac{7}{3}$
Thus $a, b, c$ are roots of the equation $P(x), x^3-5x^2+7x-\frac{7}{3}=0$.
We have,
$P(0) = -\frac{7}{3} <0$
$P(1) = 1-5+7-\frac{7}{3} >0$
$P(2) = 8-20+14-\frac{7}{3} <0$
$P(3) = 27-45+21-\frac{7}{3} >0$.
by Intermediate value theorem, $P(x)$ has a root in each of the interval $(0,1), (1, 2), (2,3)$.
Since $a<b<c$, we get $0<a<1<b<2<c<3$.

This is a classical exercise about symmetric functions. By Newton's identities
$$ e_1 = p_1 = 5, $$
$$ 2e_2 = p_1^2 - p_2 = 14, $$
$$ 3e_3 = e_2 p_1 - e_1 p_2 + p_3 = 7 $$
hence $a,b,c$ are roots of the polynomial
$$ p(z)=(z-a)(z-b)(z-c) = z^3-e_1 z^2+e_2 z-e_3 = z^3-5z^2+7z-\frac{7}{3} $$
and the claim follows from the fact that $p(0)<0, p(1)>0, p(2)<0, p(3)>0$.
Numerically the roots lie at
$ \{0.485131\ldots,1.72235\ldots,2.79252\ldots\} $.

To perform the above manipulations in a single step, one may exploit the $\text{EXP}/\text{LOG}$ map explained here and compute the Taylor series at the origin, up to the cubic term, of
$$ \exp\left(-5z+\frac{11}{2}z^2-9z^3\right)=1+5z+7z^2+\frac{7}{3}z^3+\ldots$$
to recover the coefficients of $p(z)$.
