This will give you a run for your money
a + b 3 a b + 1 + b + c 3 b c + 1 + a + c 3 a c + 1 If a , b and c are non-negative reals satisfy a b + b c + c a = 1 , find the minimum value of the expression above.
The answer is 4.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
can we solve it using convex functions. what do we call it titu's lemma. it is cauchy schwartz inequality.
Log in to reply
emmm......this titu's lemma can be deducted from cauchy schwartz inequality
when i saw this problem. first thing i wanted to do was to fix the value of a. considering the expression as a function of b,c. then we can find the minimum value of the this function. but in this problem this technique wont work.
Since these are non NEGATIVE. Therefore an expression containing only additives and multiplication can be minimised if we consider one of the variables to be 0. NEXT, USING simple AM GM inequalities, we get answer as 4
Therefore an expression containing only additives and multiplication can be minimised if we consider one of the variables to be 0
Why must this be true?
Log in to reply
I would love to post my solution, but this will give away the asnwer for Inspired by Gurido Chong
Because, for any of the values that the variable can hold, 0 is the least, so if we are able to remove one variable from the equation by automatically putting its least value and still not getting an indeterminate form then by solving, then we can achieve at its minimum value.
Log in to reply
That made no sense. Just because it sounds intuitive, it doesn't make it true.
Log in to reply
@Pi Han Goh – Oh! :( Well, one more thing can be said that, Since the minimum value of the expression, when considering all the no.s to be positive, is 3(3)^(1/2) So, as 4 is less than 3(3)^(1/2) for any particular a, b or c to be zero, for 4 to come as minimum value..
Log in to reply
@Navneel Mandal – Your explanation is not correct either. You need to prove that this must be true as well.
we can assume any particular a, b or c to be zero, for 4 to come as minimum value..
Log in to reply
@Pi Han Goh – Well, I think what @Navneel Mandal meant was when only positive reals are considered, if the expression attains a minimum of > 4 , then we know the minimum must be attained when one of a , b , c is zero. However, that does not justify the minimum as 4 . Some further work is needed to prove it. (which should be easy, just use the constraint a b + b c + a c = 1 and AM-GM).
We call the expression A, subtituting a b + b c + a c = 1 into A A = a + b + c + 4 ( a + b a b + b + c b c + a + c a c ) Now using Titu's Lemma we have A ≥ a + b + c + a b ( a + b ) + b c ( b + c ) + a c ( a + c ) 4 ( a b + b c + a c ) 2 We see that a b ( a + b ) + b c ( b + c ) + a c ( a + c ) = ( a + b + c ) ( a b + b c + a c ) − 3 a b c = a + b + c − 3 a b c ≤ a + b + c
Therefore A ≥ a + b + c + a + b + c 4 ≥ 4 So the minimum value of A is 4 and the equality holds when ( a ; b ; c ) = ( 1 ; 1 ; 0 ) and its permutations