let a,b,c be three distinct real numbers such that each of expression ax2+bx+c,bx2+cx+a,cx2+ax+b are positive for all x ε R and let

α=bc+ca+ab/a2+b2+c2 then

(A) α<4 (B) α<1 (C) α>1/4 (D) α>1



Joshi sir comment

if ax2+bx+c is positive for real x

then b2< 4ac similarly others

on adding all inequalities we get b2 + c2 + a2< 4(ac+ab+bc)

now get answer 

Read 1 Solution.

ax² + bx + c = y (let)

or, ax² + bx + (c-y) = 0

since, x is real, b² - 4a(c-y) ≥0

or, y ≥ c-b²/4a

by question, y>0

c-b²/4a >0

b² < 4ac

similarly, a² < 4bc and c² < 4ab

a² + b² + c² < 4(ab + bc+ ca)

α > 1/4

DIPANKAR GUPTA 7 year ago is this solution helpfull: 7 0

Submit Your Answer

please login to submit your answer

Login Here

Username / Email :

Password :

Register | Forget Password