For a , b , c < 1. Minimize M = a 2 ( 1...

Alex Baird

Alex Baird

Answered

2022-07-23

For a , b , c < 1. Minimize M = a 2 ( 1 2 b ) b + b 2 ( 1 2 c ) c + c 2 ( 1 2 a ) a
My Try: From a b + b c + c a = 1 3 a + b + c < 3
We have: M = a 2 b + b 2 c + c 2 a 2 ( a 2 + b 2 + c 2 )
( a + b + c ) 2 a + b + c 2 ( a 2 + b 2 + c 2 ) 4 ( a b + b c + c a ) + 4 ( a b + b c + c a )
= a + b + c 2 ( a + b + c ) 2 + 4 ( 1 )
Let a + b + c = x ( 3 x < 3 )
We find Min of function y = 2 x 2 + x + 4
And y M i n = 2 + 3 when x = 3
Right or Wrong ?

Answer & Explanation

jbacapzh

jbacapzh

Expert

2022-07-24Added 18 answers

Let a = b = c = 1 3 . Hence, M = 3 2
We'll prove that it's a minimal value.
Indeed, Let c = min { a , b , c }. Hence, we need to prove that
a 2 b + b 2 c + c 2 a 2 ( a 2 + b 2 + c 2 ) a b + a c + b c ( 3 2 ) a b + a c + b c
or
a b + a c + b c ( a 2 b + b 2 c + c 2 a 3 ( a b + a c + b c ) ) 2 ( a 2 + b 2 + c 2 a b a c b c )
or
a 2 b + b 2 a a b + b 2 c + c 2 a b 2 a c + a + b + c 3 ( a b + a c + b c )
2 ( ( a b ) 2 + ( c a ) ( c b ) )
or
( a b ) 2 ( a + b ) a b + ( c a ) ( c 2 b 2 ) a c + a + b + c 3 ( a b + a c + b c )
2 ( ( a b ) 2 + ( c a ) ( c b ) )
or
( a b ) 2 ( 1 a + 1 b 2 ) + ( c a ) ( c b ) ( 1 a + b a c 2 ) + a + b + c 3 ( a b + a c + b c ) 0 ,
which is true because 1 a + 1 b 2 > 0, 1 a + b a c 2 1 a + 1 a 2 > 0 and
a + b + c 3 ( a b + a c + b c ) = ( a b ) 2 + ( c a ) ( c b ) a + b + c + 3 ( a b + a c + b c ) 0.
Done!
The inequality
a 2 b + b 2 c + c 2 a 2 ( a 2 + b 2 + c 2 ) a b + a c + b c ( 3 2 ) a b + a c + b c
is true for all positives a, b and c, but my proof of the last statement is very ugly.

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