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.:)

The following relation:
$$(ab+bc+ca)(\frac{1}{b}+\frac{1}{c}+\frac{1}{a})\geq(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$$ is easily obtained by C-S inequality.
Since you arleady know $abc=1$ , you can easily notice that $$ab+bc+ca=\frac{1}{b}+\frac{1}{c}+\frac{1}{a}$$
and your question becomes much easier at this point.
What you should be careful about is that when $ab+bc+ca=\sqrt{a}+\sqrt{b}+\sqrt{c}$ , you get $a=b=c=1$ , which apparently doesn't fit the condition in the original question.
Therefore,
$$ab+bc+ca=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>\sqrt{a}+\sqrt{b}+\sqrt{c}$$
is proven.
