Find the least value of :-
(a+b+c)(1/a+1/b+1/c)
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Answered by
4
Given that values of a, b and c are real and independent.
The given expression Z = (a+b+c) (1/a + 1/b + 1/c) is symmetric in a , b and c.
So at the extremum of maximum or minimum, a = b = c.
Z = (3a)(3/a) = 9.
Min value is 9.
Clearly the maximum value is infinity. We can choose a and b to be very large. c can be chosen as 1. Then the second factor is nearly 1. The first factor is very large.
So the minimum is 9.
The given expression Z = (a+b+c) (1/a + 1/b + 1/c) is symmetric in a , b and c.
So at the extremum of maximum or minimum, a = b = c.
Z = (3a)(3/a) = 9.
Min value is 9.
Clearly the maximum value is infinity. We can choose a and b to be very large. c can be chosen as 1. Then the second factor is nearly 1. The first factor is very large.
So the minimum is 9.
Anonymous:
thanks sir
.★.
It is not clear if a, b, and c are sides of a triangle... or all independent numbers. Assuming they are positive real numbers...
Take some positive integers/real values .. check if 9 is the minimum value by substituting the values numerically
you couldTake some positive integers/real values .. check if 9 is the minimum value by substituting the values numerically
√.
independent variables
numbers*
we could substitute a = 3, b = 49.4 , c = 2 .. like this and check, if result is more than 9 always
okay !
Answered by
3
Heya user,
[ a + b + c ] [ 1/a + 1/b + 1/c ]
= [ a + b + c ] [ 1² / a + 1² / b + 1² / c ]
Applying Titu's lemma, this changes to :->
[ a + b + c ] [ 1/a + 1/b + 1/c ] >= [ a + b + c ] [ 1 + 1 + 1 ]² / [ a + b + c ] = 3² = 9;
=> [ a + b + c ] [ 1/a + 1/b + 1/c ] >= 9;
Hence, the minimum value is 9;
[ a + b + c ] [ 1/a + 1/b + 1/c ]
= [ a + b + c ] [ 1² / a + 1² / b + 1² / c ]
Applying Titu's lemma, this changes to :->
[ a + b + c ] [ 1/a + 1/b + 1/c ] >= [ a + b + c ] [ 1 + 1 + 1 ]² / [ a + b + c ] = 3² = 9;
=> [ a + b + c ] [ 1/a + 1/b + 1/c ] >= 9;
Hence, the minimum value is 9;
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