Nice Use of?
1 − a a ⋅ 1 − b b ⋅ 1 − c c
For a , b , c are real numbers such that 0 < a , b , c < 1 with a + b + c = 2 . Find the minimum value of the above expression.
The answer is 8.
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3 solutions
Moderator note:
This is not a solution.
Can you explain how you get a=b=c using AM-GM inequality...
thanks.. ......
This is not a solution. You have not bounded the expression.
Are you sure that ∑ a 1 − a is maximized at a = b = c ? What happens as a → 0 ?
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I don't understand, to be honest. As a becomes smaller, b+c has to get bigger. Right? Let us say that a=0,01
Then at least one of b and c has to be 0,995 and larger, and none of them can be less than 0,99. When trying this on my calculator it seems like the expression decreases as a does so. Am I thinking completely wrong here?
Also, could you, Sir, explain to me shortly whats wrong with my solution? Is it the inverting part that is not allowed to do, or can I not use AM-GM on this expression(sure, from the problem formulation it follows that all terms are positive, but without those limits they might as well be negative.. )
Thank you
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So, what you have shown, is that 3 3 ∏ a 1 − a ≤ ∑ a 1 − a . I agree with this statement.
However, the RHS is not bounded above, as seen with a → 0 . For example, with a = 1 0 − n ( and b = c = 1 − 2 1 1 0 − n , but that doesn't matter), we get a 1 − a ≈ a 1 = 1 0 n .
So, you don't have an upper bound on the LHS, which means that you don't have a (non-zero) lower bound on the inverse.
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@Calvin Lin – Thank you for telling me! I will read some papers about AM-GM for grasping, realize now that I have been doing stupid assumptions(AM-GM just lost half it's power, for me). However, my inverting idea, is it possible to do?
When the RHS or LHS is not fixed, one cannot assume that equality means maximum respectively minimum?
Have a great day
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@Adam Pet – Your inverting idea is valid (but I don't know if it will lead to a solution). The issue is whether or not you can prove that there is a maximum.
1 − a a ⋅ 1 − b b ⋅ 1 − c c = a , b , c ∏ ( 1 − a 1 − 1 ) = − 1 + ∑ 1 − a 1 − c y c ∑ 1 − a 1 1 − b 1 − ∏ 1 − a 1 = − 1 + ∑ 1 − a 1 ( ∵ a + b + c = 2 ) ≥ − 1 + ∏ ( 1 − a ) 1 / 3 3 ≥ − 1 + 9 = 8 the last two steps follow from AM-GM inequality.
For the minimum value, take a = b = c
So we get a = b = c = 3 2 = . 6 6
Thus 1 − a a × 1 − b b × 1 − c c
= . 3 3 . 6 6 × . 3 3 . 6 6 × . 3 3 . 6 6 = 2 × 2 × 2 = 8
Moderator note:
This is not a solution.
what is the reason for taking a=b=c?
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How do you Solve ?
@Parth Lohomi Please do not set all of your problems as level 5. If this continues, I might be forced to make you unable to set problems at level 5.
For most of the questions asking minimum value, try this method
DAMN !
We notice that the minimum value of the given expression is given by the same solution as for the maximum for the inverted equation.
Using AM-GM, one get that a=b=c gives the maximum value of the inverted expression. Hence, a=b=c also gives the minimum of the original expression.
From the sum of a + b+ c, we see that a=b=c=2/3, which makes each factor 2, and the product is therefore 8.
The Left side is the inverted expression. The values of a,b,c that gives the maximum for this, will also give the minimum to the original.
From the AM-GM, we now that equality occurs when the three terms/factors are the same. Try this, and you will see that then a=b=c.
Correct?