If $a,b,c\in\mathbb{R^+}$ such that $ abc = 1 $ and $ ab + bc + ca = 5 $. Prove that $ 17/4 \leq (a+b+c)\leq 1+ \sqrt{32}. $ If $a,b,c\in\mathbb{R^+}$ such that $ abc = 1 $ and $ ab + bc + ca = 5 $. Prove that 
$$ 
\frac{17}{4} \leq (a+b+c)\leq 1+ \sqrt{32}. 
$$
My attempt
Tried using Vieta but it didn't work. Also I  used some standard inequalities but got $ \geq\sqrt{15} $.
 A: Let $P(x)=(x-a)(x-b)(x-c)$. 
$$P(x)=x^3-sx^2+5x-1$$ 
with $s=a+b+c$
$P$ must have 3 reals roots. So discrimant of $P$ must be positive. From the formula of discriminant, we get
$$\Delta(P)=-4s^3+25s^2+90s-527=R(s)$$
But the polynomial $R$ has an easy root : $\frac{17}{4}$. So we factorize :
$$R(s)=-4\left(s-\frac{17}{4}\right)(s^2-2s-31)=-4\left(s-\frac{17}{4}\right)\left(s-(1+\sqrt{32})\right)\left(s-(1-\sqrt{32})\right)$$
So $R$ is only positive between its two positive roots, so
 $$\frac{17}{4}\le s\le 1+\sqrt{32}$$
Note : $R$ is also positive before its negative root, but it would imply $s<0$.

Another solution : The limit cases where $P$ has 3 real roots are when $P$ has two identical roots. So just suppose $b=c$.
Hence $ab^2=1$ and $2ab+b^2=5$ so $Q(b)=b^3-5b+2=0$ 
But $Q$ has some easy roots : $2$, and $-1\pm\sqrt{2}$. 
Use the two positive roots as values for $b$, and deduce $a$ and $a+b+c$, you will find again $\frac{17}{4}$ and $1+\sqrt{32}$.
A: The following approach does not make use of the theory of third degree equations. At the same time it explicitly gives the extremal triples.
If $(a,b,c)$ is an admissible triple one also has
$${1\over a}+{1\over b}+{1\over c}=5\ .$$
This implies that $a$, $b$, $c$ all are $>{1\over5}$, and as $abc=1$ all have to be $<25$. It follows that the admissible triples form a compact set $K\subset\bigl[{1\over5},25\bigr]^3$, and the quantity $s:=a+b+c$ assumes finite extremal values on $K$. At the end of this post I shall prove the following 
Lemma. At a point $(a,b,c)\in K$ where $s$ is extremal at least two of the three numbers $a$, $b$, $c$ have to be equal.
Therefore it is sufficient to consider triples $(a,a,b)\in K$. For these one has
$$a^2 +2ab=5, \qquad a^2b=1\ .$$
This implies $a^3-5a+2=0$ with the solutions
$$2,\quad\sqrt{2}-1,\quad-\sqrt{2}-1\ .$$
Therefore we obtain the two admissible triples
$$\left(2,2,{1\over4}\right),\qquad\bigl(\sqrt{2}-1,\>\sqrt{2}-1,\>3+2\sqrt{2}\bigr)\ ,$$
which are bound to produce the extremal values for $s$. One easily verifies that we indeed have
$${17\over4}\leq s\leq1+4\sqrt{2}\ .$$
Proof of the Lemma. The map
$$\Phi:\quad(a,b,c)\mapsto(a+b+c,\>ab+bc+ca, \>abc)$$
has Jacobian
$$J_\Phi(a,b,c)=(b-a)(c-b)(a-c)\ .$$
If $(a_0,b_0,c_0)$ is an admissible triple with $a_0$, $b_0$, $c_0$ all different then $J_\Phi(a_0,b_0,c_0)\ne0$. It follows that the restriction of $\Phi$ to a suitable neighborhood $U$ of $(a_0,b_0,c_0)$ maps $U$ bijectively onto a neighborhood $V$ of $\Phi(a_0,b_0,c_0)=(s_0,5,1)$, where  $s_0:=a_0+b_0+c_0$. Therefore the inverse map $\Psi:\> V\to U$ can be used for the function $$\psi:\quad s\mapsto \Psi(s,5,1)\in K\qquad(s_0-h<s<s_0+h)\ ,$$
which produces admissible triples whose $s$-value is $>s_0$ and other admissible triples whose $s$-value is $<s_0$.
A: Let $p=a+b+c,\,q=ab+bc+ca=5,\,r=abc=1$ we have
$$\begin{aligned}0 \leqslant (a-b)^2(b-c)^2(c-a)^2 & = -27r^2-2p(2p^2-9q)r+q^2(p^2-4q) \\&= (17-4p)(p^2-2p-31).\end{aligned}$$
Now, solve inequalities
$$(17-4p)(p^2-2p-31) \geqslant 0,$$
we get
$$\frac{17}{4} \leqslant p \leqslant 1 + 4 \sqrt 2.$$
The proof is completed.
A: In order that the polynomial
$$ p(z) = z^3 - sz^2 + 5z-1 $$
has three real roots we need that it is not an increasing function, so the discriminant of
$$ p'(z) = 3z^2 - 2sz + 5 $$
has to be non negative, from which $|s|\geq \frac{1}{2}\sqrt{15}$ as you stated. Since we want the three roots to be positive, $s>0$. If $p(z)$ has three positive roots, $p'(z)$ has two positive roots $\xi_1,\xi_2$ with $\xi_1<\xi_2$ and $p(\xi_1)>0$ while $p(\xi_2)<0$. Now, just write down the last two conditions.
A: WLOG, using the symmetry, we may assume $a \le b \le c$, so $a \le 1, \; c \ge 1$.
For the left inequality, note that
$$a+b+c-\frac{17}4 = \frac{5-1/c}{c}+c-\frac{17}4 = \frac{(c-2)^2(4c-1)}{4c^2} \ge 0$$
and as equality is achieved when $a=\frac14, b=c=2$ (or any permutation), this in fact gives the minimum. 
Similarly, for the right inequality, note
$$a+b+c-(1+4\sqrt2) = a+\frac{5-1/a}{a}-(1+4\sqrt2) \\ = -\frac{(a+1-\sqrt2)^2(3+2\sqrt2-a)}{a^2} \le 0$$
and as equality is possible when $a=b=\sqrt2-1, c= 3+2\sqrt2$, this gives the maximum.

P.S. For a bonus, the above expressions also indicate that $\min(a, b, c) \ge \frac14$ and $\max(a, b, c) \le 3+\sqrt2$ (why?).
