A Nice Inequality
2 a + b 1 + 2 b + c 1 + 2 c + a 1 If a , b and c are positive reals satisfying a 1 + b 1 + c 1 = 1 , find the maximum value of the expression above.
Submit your answer to 3 decimal places
The answer is 0.333.
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3 solutions
How did you arrive at the first inequality?
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By applying Cauchy-Schwarz Inequality
How did you use Cauchy-Schwarz in the first line? Could you explain in full?
a + b + c ≥ 9 and a + b + c ≥ 3 3 a b c doesn't make a b c = 2 7
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From (2) and AM-GM inequality,minimum of a+b+c is 9=3 x abc^(1/3) .Can you explain in what light you are trying to say that?
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try a = 1 9 3 0 ; b = 6 ; c = 5 , then a + b + c ≥ 9 and a + b + c ≥ 3 3 a b c still holds true but a b c = 2 7 is wrong. By saying a b c = 2 7 , you are assuming a = b = c
By Titu's Lemma ,
a 1 + a 1 + b 1 ≥ 2 a + b 9
b 1 + b 1 + c 1 ≥ 2 b + c 9
c 1 + c 1 + a 1 ≥ 2 c + a 9
Adding all gives ,
a 3 + b 3 + c 3 ≥ 2 a + b 9 + 2 b + c 9 + 2 c + a 9
⇒ 9 3 = 3 1 ≥ 2 a + b 1 + 2 b + c 1 + 2 c + a 1
Equality holds when a = b = c = 3
Call the expression M we have 2 a + b 9 ≤ a 1 + a + b 4 ( T i t u ′ s L e m m a ) Doing the similar to the others then combine 3 inequalities, we have 9 M ≤ 1 + a + b 4 + b + c 4 + a + c 4 ≤ 3 ( A M − H M ) ⇔ M ≤ 3 1 The equality holds when a = b = c = 3