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.:)
 A: 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.
A: By your work $abc=1$.
Let $a=x^2$, $b=y^2$ and $c=z^2$, where $x$, $y$ and $z$ are positives.
Thus, $$xyz=1$$ and we need to prove that:
$$\sum_{cyc}\frac{1}{x^2}>x+y+z$$ or
$$\sum_{cyc}(x^2y^2-x^2yz)>0$$ or
$$\sum_{cyc}z^2(x-y)^2>0,$$
which is true because $x=y=z$ is impossible.
Indeed, let $x=y=z$.
Thus, $a=b=c=1$ and $R=\frac{1}{\sqrt3},$ which is a contradiction.
A: $ \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.
A: 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
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}> \sqrt a+ \sqrt b + \sqrt c$
