11AM-GM for 1st sum....
Ques1) If a > 0 , b > 0 , c > 0 and the min value of a (b2 + c2) + b(c2 +a2 ) + c(a2+b2) is k abc then prove that k =6.
Ques2) If a,b,c are three real numbers such that b+c-a , c+a-b , and a+b-c are positive, then the expression (b+c-a) (C+a-b) -abc is
(a)positive (b) negative (c) non positive (d) non negative
Ques3) If a,b,c are in A.P ; p,q,r in H.P and ap,bq,cr are in G.P. then show that p/r + r/p is aquals to a/c + c/a.
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3 Answers
62First one directly apply AM GM on ab^2, ac^2, bc^2.. and so on.. (motivation.. you have a 6 .. so there could be 6 terms!)
Third one..
a+c=2b
1/p+1/r=2/q
(p+r)q/2=pr
apcr=b^2q^2
(p2+r2)/rp =(p2+r2)/rp+2-2
=(p+r)2)/rp - 2
=4p2r2q2(rp) -2
=4prq2 -2
= 4b2/ac - 2
= (a+c)2/ac - 2
= (a2+c2)/ac
= a/c + c/a
66For the first one, we have
a(b2+c2)+b(c2+a2)+c(a2+b2)abc = bc + cb + ca + ac + ab +ba ≥ 6
