another inequality 1 x y + z + 1 y z + x + 1...

Willow Pratt

Willow Pratt

Answered

2022-07-05

another inequality 1 x y + z + 1 y z + x + 1 z x + y 1 2
Let x , y , z > 0 and such x y z 2 + x + y + z, show that
(1) 1 x y + z + 1 y z + x + 1 z x + y 1 2
The theory basis of speculation Use the following classical results
x y z = 2 + x + y + z x y + y z + z x 2 ( x + y + z )

Answer & Explanation

Kaya Kemp

Kaya Kemp

Expert

2022-07-06Added 18 answers

I agree with you. Your inequality is true!
Indeed, the condition gives c y c 1 x + 1 1
Let x = a, y = b and z = k c, where k > 0 and c y c 1 a + 1 = 1. Hence,
1 a + 1 + 1 b + 1 + 1 c + 1 1 a + 1 + 1 b + 1 + 1 k c + 1
which gives k 1
Thus, c y c 1 x y + z = 1 a b + k c + 1 k a c + b + 1 k b c + a 1 a b + c + 1 a c + b + 1 b c + a
Id est, it remains to prove that c y c 1 a b + c 1 2 for positives a, b and c such that c y c 1 a + 1 = 1
Let a = y + z x and b = x + z y , where x, y and z are positive numbers.
Hence, c = x + y z and we need to prove that c y c 1 x + y z x + z y + y + z x 1 2 or
c y c x y z x ( x + y ) ( x + z ) + y z ( y + z ) 1 2
or
c y c x y z ( x + y + z ) ( x 2 + y z ) 1 2
or
c y c ( x 2 x y z x 2 + y z ) 0
or
c y c x ( x 2 y z ) x 2 + y z 0
or
c y c x ( ( x y ) ( x + z ) ( z x ) ( x + y ) ) x 2 + y z 0
or
c y c ( x y ) ( x ( x + z ) x 2 + y z y ( y + z ) y 2 + x z ) 0
or
c y c z ( x y ) 2 ( x 2 + y 2 + x z + y z ) ( z 2 + x y ) 0
Done!

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