<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] 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.
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