Given positive reals a , b , c such that a + b + c = 4 , what is the maximum value of
a 2 b c + a b 2 c + a b c 2 ?
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but this is giving maximum value. .you should have asked maximum value
By A.M.-G.M. inequality a+b+c / 3 ≥ (abc)^1/3
But given a+b+c=4
therefore a+b+c / 3 ≥ (abc)^1/3 =>4/3 ≥ (abc)^1/3 =>64/27 ≥ abc ---->(1)
Now a^2bc+b^2ac+c^2ab can be written as abc(a+b+c) =>4abc ---->(2)
(1)=>The largest value of abc is 16/27. By substituting abc in (2)
We get abc(a+b+c)</= 4 x 64/27=256/7
therefore the maximum value of a^2bc+b^2ac+c^2ab is 256/7
The above solution gives the maximum value.Please edit the question
it is >= means min 256/7 or more than that therefore min value only baby.......
Can you fix the inequality in the third line? It should be 6 4 / 2 7 ≥ a b c . please fix the follow through errors too.
Can you fix the inequality sign and the denominator in the 3rd last line? It should be
a b c ( a + b + c ) ≤ 4 × 6 4 / 2 7 = 2 5 6 / 2 7
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Apply AM-GM inequality on a , b , c 3 a + b + c ≥ 3 a b c 4 / 3 ≥ 3 a b c 2 7 6 4 ≥ a b c Therefore for maximum value take a b c = 2 7 6 4 a 2 b c + b 2 c a + c 2 a b = a b c ( a + b + c ) = 2 7 6 4 × 4 = 2 7 2 5 6