<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] 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?).
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