<s>[INST] 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} $.
 [/INST] 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 $1+\sqrt{32}$.
</s>