An algebra problem by A Former Brilliant Member
Find the minimum value of a a b b c c , where a , b , and c are positive real numbers such that a + b + c = 1 .
The answer is 0.333.
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2 solutions
Elegant and compact solution.
Just enclose your formulas with "\ (" and "\ )" (without spaces between backslash "\" and brackets "()") to effect the LaTex code.
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I'm too lazy to use LaTex at this age :D
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I am 61. You need only 6 more keystrokes. I amended your problem I added 12 keystrokes for Latex. I have rephrased it to read better. It is unnatural to start a sentence in lowercase.
Which sentence? If it was so, I'm very sorry for that typing error.
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You original problem starts with "a, b, c are positive integers...".
Because of symmetry, a==b==c, Using the constraint a+b+c=1 and replacing b and c with a gives 3a=1 or a = 3 1 . Evaluating a a b b c c as 3 1 1 gives the minimum as 3 1 .
Your premise is not always correct. See this article .
After reviewing the article and concluding the argument had sufficient validity to warrant a reworking of this problem from the start, I did just that.
I plotted
a
a
between 0 and 1:
To find the minimum of a a , ∂ a ∂ a a ⇒ a a ( lo g ( a ) + 1 ) which, since the function is concave upwards, means that the minimum is at e 1 .
This can be used as an estimate of where the minimum is. The minimum can not be there exactly as e 3 > 1 .
I plotted the expression in an appropriate range, noting that because if the
a
+
b
+
c
=
1
constraint, the valid region will be triangular by the origin:
This supports my original solution.
Let us see what Wolfram Mathematica 12 has to say, Assuming [ ( a ∣ b ) ∈ R , Minimize [ { a a b b ( 1 − ( a + b ) ) 1 − ( a + b ) , 1 0 2 ≤ a ≤ 1 0 4 ∧ 1 0 2 ≤ b ≤ 1 0 4 } , { a , b } ] ] ⇒ { 3 1 , { a → 3 1 , b → 3 1 } }
Vindicated.
We note that f ( x ) = x ln x is convex. Hence by Jensen's inequality , we have:
3 a ln a + b ln b + c ln c a ln a + b ln b + c ln c ⟹ a a b b c c ≥ 3 a + b + c ln ( 3 a + b + c ) ≥ − 3 ln 3 ≥ − ln 3 ≥ 3 1 ≈ 0 . 3 3 3 And equality occurs when a = b = c = 3 1