How prove this inequality $\dfrac{R}{r}\ge\dfrac{b}{c}+\dfrac{c}{b}$ in $\Delta ABC$,prove that
$$\dfrac{R}{r}\ge\dfrac{b}{c}+\dfrac{c}{b}$$
where $R$ is the circumradius and $r$ is the inradius
By the way.It is well konwn that Eluer inequality
$$R\ge 2r$$
and it is easy proof: by $$d^2=R(R-2r)$$
see http://mathworld.wolfram.com/EulersInequality.html
and  $$ \dfrac{R}{r}\ge\dfrac{b}{c}+\dfrac{c}{b}$$
is shaper than Eluer inequality, I try solve it,But I can't work, But after I can find This is 
famous inequality:see 
http://books.google.com.hk/books?id=hXYH2OfNRdwC&pg=PA177&lpg=PA177&dq=Gaz.+Mat.+(Bucharest)+90+(1985),+65&source=bl&ots=xP0JX0PNta&sig=RdcaDooKrw-0wvdVsj4C35E5AKY&hl=zh-CN&sa=X&ei=qXgHUsOUHMWziQeT8IG4DA&ved=0CC4Q6AEwAA#v=onepage&q=Gaz.%20Mat.%20(Bucharest)%2090%20(1985)%2C%2065&f=false
(5.30),But I can't find this equality solution too,Thank you someone can find it or take this inequality methods。Thank you everyone.
 A: $l_{a}$ is $A$-inner angle bisector, where $l_{a}^{2}=\dfrac{4bc\cdot s(s-a)}{(b+c)^{2}}\leq s(s-a)$
$\therefore h_{b}^{2}+h_{c}^{2}\leq l_{b}^{2}+l_{c}^{2}\leq s(s-b)+s(s-c)=sa$, where $h_{b}$ is $B$-altitude    
$\Longrightarrow$ $h_{b}^{2}+h_{c}^{2}=4(\triangle ABC)^{2}\cdot\left(\dfrac{1}{b^{2}}+\dfrac{1}{c^{2}}\right)=\dfrac{abc}{R}\cdot rs\cdot \left(\dfrac{1}{b^{2}}+\dfrac{1}{c^{2}}\right)\leq sa$
$\therefore$ $\dfrac{R}{r}\geq \dfrac{b}{c}+\dfrac{c}{b}$
A: en, Notice that
$$S=\frac{abc}{4R}, S=pr$$
where $p=\frac{a+b+c}2$.Heron's formula give $S=\sqrt{p(p-a)(p-b)(p-c)}$, we get that
 $$\frac Rr=\frac{2abc}{(a+b-c)(b+c-a)(c+a-b)}$$
then, take $a=y+z,b=z+x,c=x+y$, yields 
$$\frac Rr=\frac{(x+y)(y+z)(z+x)}{4xyz}$$
then, 
$$\frac bc+\frac cb=\frac{z+x}{x+y}+\frac{x+y}{z+x}$$
so, it is sufficient to show that 
$$(x+y)^2(y+z)(z+x)^2\geq 4xyz[(z+x)^2+(x+y)^2]$$
ha, $(x+y)^2z\geq4xyz$ give
$$z(x+y)^2(z+x)^2\geq 4xyz(z+x)^2,$$
and $y(z+x)^2\geq4xyz$ yields
$$y(x+y)^2(z+x)^2\geq 4xyz(x+y)^2$$
we get the result
