Let a,b and c be the side lengths of a triangle ABC and assume that a≤b and a≤c. If x=b+c−a2, then the minimum value of axrR, where r and R denote the inradius and circumradius, respectively of triangle ABC, is
Open in App
Solution
Let y=axrR=a(b+c−a)2⋅△s⋅abc4△ =a(b+c−a)(b+c+a)abc =(b+c)2−a2bc ⇒y=(bc+cb+2)−a⋅abc
As a≤b and a≤c ⇒ab≤1 and ac≤1 ∴a2bc≤1 ∴y≥(bc+cb+2)−1
Using A.M. ≥ G.M., bc+cb≥2 ⇒y≥2+2−1 or y≥3 ⇒ymin=3 (equality holds when a=b=c)