How to prove the following inequality of logarithm? Let x , y , z &#x2208;<!-- ∈

Fletcher Hays

Fletcher Hays

Answered question

2022-06-24

How to prove the following inequality of logarithm?
Let x , y , z C . Suppose
z = 1 2 ( x y ± x 2 y 2 4 ( x 2 + y 2 ) ) .
Show that
l o g + | z | l o g + | x | + l o g + | y | + l o g 2.
Where l o g + ϕ = m a x { 0 , l o g ϕ } .
Here we are also considering the complex root function with respect to the principle branch of logarithm.

Answer & Explanation

Jovan Wong

Jovan Wong

Beginner2022-06-25Added 23 answers

To prove what you want, we basically just need to apply the triangle inequality to your formula. I'll look at cases separately depending whether | x | , | y | are greater than or smaller than unit magnitude.
1. If | x | 1, | y | 1, then | x 2 y 2 4 x 2 4 y 2 | | x 2 y 2 | + | 4 x 2 | + | 4 y 2 | 9 so
2 z = x y ± ( x 2 y 2 4 x 2 4 y 2 ) 1 / 2 ,
| 2 z | | x y | + | x 2 y 2 4 x 2 4 y 2 | 1 / 2 1 + 9 1 / 2 = 4
Hence log | z | log 2
2. If | x | 1, | y | 1, then
z x = y 2 ± ( y 2 4 y 2 x 2 1 ) 1 / 2 ,
| z x | | y | 2 + | y 2 4 y 2 x 2 1 | 1 / 2 | y | 2 + ( | y | 2 4 + | y 2 x 2 | + 1 ) 1 / 2 1 2 + | 1 4 + 2 | 1 / 2 = 1 2 + | 9 4 | 1 / 2 = 2
Hence log | z x | log 2
3. If | y | 1, | x | 1, then by symmetry with x and y and case 2 we have log | z y | log 2
4. If | x | 1, | y | 1, then | 1 / x 2 + 1 / y 2 | 2 so,
z x y = 1 2 ± ( 1 4 1 x 2 1 y 2 ) 1 / 2 ,
| z x y | 1 2 + | 1 4 1 x 2 1 y 2 | 1 / 2 1 2 + ( 1 4 + | 1 x 2 + 1 y 2 | ) 1 / 2 1 2 + | 1 4 + 2 | 1 / 2 = 1 2 + | 9 4 | 1 / 2 = 2
Hence log | z x y | log 2
In summary,
log | z | { log 2  if  | x | 1 , | y | 1 log | x | + log 2  if  | x | 1 , | y | 1 log | y | + log 2  if  | x | 1 , | y | 1 log | x | + log | y | + log 2  if  | x | 1 , | y | 1 .
Then since log + | x | = { 0  if  | x | 1 log | x |  if  | x | 1 , this is equivalent to:
log | z | log + | x | + log + | y | + log 2.

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