<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 $p=a+b+c,\,q=ab+bc+ca=5,\,r=abc=1$ 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.
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