IIT JEE , AIEEE Forum - Mathematics , Algebra - Question from A.M

btetarbe

If a,b,c are real and distinct then show that:
a²(1 + b²) + b²(1 + c²) + c²(1 + a²) > 6abc

sboosy

a²(1+b²)+b²(1+c²)+c²(1+a²) =

a²+b²+c²+a²b²+b²c²+c²a²

now AM>=GM for the first three

(a²+b²+c²)/3 >= ^{[3 ]}

a²b²c²that is abc whole power 2/3

similarly (a²b²+b²c²+c²a²)/3>= abc whole power 4/3

taking 3 to the other side in both the above

the given >= 3 (abc)power 1/3 [ abc + (abc power1/3) ]

now applying am>=gm for abc + abc power1/3

the value is greater than or equal to 2 abc power 2/3

so the given >= 3 abc(power 1/3) * 2 abc(power 2/3)

>=6abc

question to b corrected

there is equality

priyesh

a²(1+b²)+b²(1+c²)+c²(1+a²) =

a²+b²+c²+a²b²+b²c²+c²a²

applying A.M >= G.M

(a²+b²+c²+a²b²+b²c²+c²a²)/6>= (a⁶b⁶c⁶)^1/6 = abc

h.p

sboosy

nice one my friend