Let (x,y,z) in R, prove that ((2x-y)/(x-y))^2+((2y-z)/(y-z))^2 +((2z-x)/(z-x))^2 >= 5

Kelton Bailey 2022-09-30 Answered
Let ( x , y , z ) R , prove that ( 2 x y x y ) 2 + ( 2 y z y z ) 2 + ( 2 z x z x ) 2 5
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Answers (2)

smh3402en
Answered 2022-10-01 Author has 11 answers
Start with
( 2 y 2 z + 2 z 2 x + 2 x 2 y y z 2 z x 2 x y 2 3 x y z ) 2 0
This can be expanded out to give
4 c y c y 2 z 4 4 c y c y 3 z 3 + c y c y 4 z 2 4 c y c x y z 4 + 14 c y c x y 2 z 3 10 c y c x y 3 z 2 3 x 2 y 2 z 2 0
This can be rearranged to
( 2 x y ) 2 ( y z ) 2 ( z x ) 2 + ( x y ) 2 ( 2 y z ) 2 ( z x ) 2 + ( x y ) 2 ( y z ) 2 ( 2 z x ) 2 5 ( x y ) 2 ( y z ) 2 ( z x ) 2
Now divide by
( x y ) 2 ( y z ) 2 ( z x ) 2 and we have
( 2 x y ) 2 ( x y ) 2 + ( 2 y z ) 2 ( y z ) 2 + ( 2 z x ) 2 ( z x ) 2 5.
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flatantsmu
Answered 2022-10-02 Author has 1 answers
This is a special case of the following inequality:
a , b , c ( m a n b a b ) 2 m 2 + n 2 ,
where m , n R and a , b , c are distinct reals.
The proof is simply completing the square: start by writing
m a n b a b = m + ( m n ) b a b
and
m c n a c a = n + ( m n ) c c a
and leave the middle term untouched first.
Let me know if still need further assistance.
EDIT: Completing the hint:
a , b , c ( m a n b a b ) 2 m 2 n 2 = ( m b n c b c ) 2 + ( m n ) 2 ( b 2 ( a b ) 2 + c 2 ( c a ) 2 ) + 2 ( m n ) ( m b a b + n c c a ) =
= ( m b n c b c ) 2 + ( m n ) 2 ( b 2 ( a b ) 2 + c 2 ( c a ) 2 ) + 2 ( m n ) b c ( m n ) a ( m b n c ) ( a b ) ( c a ) =
= ( m b n c b c ) 2 + ( m n ) 2 ( b 2 ( a b ) 2 + c 2 ( c a ) 2 + 2 b c ( a b ) ( c a ) ) 2 a ( m b n c ) ( m n ) ( a b ) ( c a ) =
= ( m b n c b c ) 2 + ( m n ) 2 ( b a b + c c a ) 2 2 a ( m n ) ( m b n c ) ( a b ) ( c a ) =
( m b n c b c ) 2 + ( m n ) 2 ( a ( c b ) ( a b ) ( c a ) ) 2 2 a ( m n ) ( m b n c ) ( a b ) ( c a ) =
= ( m b n c b c a ( m n ) ( c b ) ( a b ) ( c a ) ) 2 0.
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