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.:)

Area of triangle ABC $=\frac{abc}{4R}=0.25$ where $R=1$ and a>0, b>0, c>0
$abc=1$
$\frac{1}{a}+\frac{1}{b}\geq\frac{2}{\sqrt{ab}}=\sqrt c$
$\frac{1}{b}+\frac{1}{c}\geq\frac{2}{\sqrt{bc}}=\sqrt a$
$\frac{1}{a}+\frac{1}{c}\geq\frac{2}{\sqrt{ac}}=\sqrt a$
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \sqrt a+ \sqrt b + \sqrt c$
Since equality holds when $a=b=c=1$ but Area of triangle $ABC=\frac{\sqrt3}{4}\not=0.25$
Then
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}> \sqrt a+ \sqrt b + \sqrt c$
