<s>[INST] To Prove $\frac{1}{b}+\frac{1}{c}+\frac{1}{a} > \sqrt{a}+\sqrt{b}+\sqrt{c}$ The three sides of a triangle are $a,b,c$, the area of the triangle is $0.25$, the radius of the circumcircle is $1$.
Prove that $1/b+1/c+1/a > \sqrt{a}+\sqrt{b}+\sqrt{c}$.
what I've tried:
$$\frac{1}{4} = \frac{1}{2}ab\sin C \Rightarrow ab=\frac{1}{2}\sin C \\c=2\sin C \Rightarrow \frac{1}{c}=\frac{1}{2}*\sin C $$
so, $$\frac{1}{c}=ab \Rightarrow abc=1 \Rightarrow \sqrt{abc}=1$$
now the problem becomes
$$ab+bc+ac > \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}},$$
with $0<a\leq b\leq c\leq 2$, and $a+b>c$.
But even so I don't know how to prove it.
Any help or hint is appreciated. Thank you.:)
 [/INST] $ \sum_{cyc}\frac{1}{a} = \frac{1}{2} \sum_{cyc}\frac{1}{a} + \frac{1}{b} \ge  \sum_{cyc}\frac{1}{\sqrt {ab}} = \sum_{cyc}\sqrt {a} \,$ (using AM-GM and $abc = 1)$
EDIT: I think I missed an important point earlier that when $a = b = c = 1$, R = $\frac{1}{2\sin 60^0} = \frac{1}{\sqrt3}$ which does not meet the condition. Hence the equality is not possible.
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