How prove this inequality $\dfrac{R}{r}\ge\dfrac{b}{c}+\dfrac{c}{b}$

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BThank you everyone.