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let $x,y$ and $z$ be complex numbers and $t\in (0,1)$ such that $y = tx + (1-t)z$ Prove that $${|z| - |y| \over |z- y|} \ge {|z| - |x| \over |z- x|} \ge {|y| - |x| \over |y- x|}$$

$$y = tx + z - zt \implies z - y = t(z - x) \implies |z- y| \le |z-x| \implies {1 \over |x -z|} \le {1 \over |z-y|} \tag{1}$$

Adding $|z-y| \ge |z| - |y|$ and $-|z-x| \ge -|z| + |x|$

$$|z-y| - |z- x| \ge |x| - |y|$$

LHS of this inequalty is $\ge 0$

$$\therefore |x| \le |y| \implies -|x| \ge -|y| \implies |z| - |y| \ge |z| - |x| \tag{2}$$

Using the original equation,

$$|y| = |tx + z(1-t)| \le t|x| + (1-t)|z| \le |x| + |z| \le |z| \tag{3}$$

$$\therefore |y| - |x| \le |z| - |x| \tag{4}$$

From $(2), (3)$

$$|z| \ge |y| \ge |x|\tag{5}$$

From $(1), (2)$ and $(5)$ ,

$${|z| - |x| \over |x -z|} \le {|z| - |x| \over |z-y|} \le {|z| - |y|\over |z-y|}$$

$$\bbox[5px, Border:2px solid black]{{|z| - |x| \over |z-y|} \le {|z| - |y|\over |z-y|}} \tag{6}$$

Now for the second part of the inequality,

$$y - x = tz + z - x - zt \implies y -x = (1-t)(z-x) \implies |y-x| \le |z-x| \implies {1\over |z-x|} \le {1\over |y-x|} \tag{7}$$ From $(4)$ and $(7)$,

$${|y| - |x|\over |y-x|} \le{|z| - |x|\over |y-x|} = {|z| - |x|\over (1-t)|z-x|} \color{red}{\ge}{|z| - |x|\over |z-x|}$$

Now I am unable to get the second part.

  • Please correct me and don't show alternate method to prove this inequality as this was one of the solved problems.
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    parameter $t$ is a real? – 2017-02-13
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    @Hassan Yes $t\in (0,1)$. – 2017-02-13
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    (5) is wrong obviously. – 2017-02-14
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    @TakahiroWaki Why ? I got $|z| \ge |y|$ from (3) and $|x| \le |y|$ from (2). Now this is simple maths $|z| \ge |y| \ge |x|$ which is (5). – 2017-02-14
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    In between $(1)$ and $(2)$ this is wrong: $$-|z-x| \ge -|z| + |x|$$ In fact, the opposite inequality holds true. – 2017-02-14
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    @dxiv Oh god you are correct. But I still first part correct, there has to be one more mistake. – 2017-02-14
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    @A---B No, there is not necessarily another mistake in the first part. Once you presume a false premise (after the first mistake), you can "prove" anything, including statements that *happen* to be true. Only that's no longer a proof after the first mistake, of course. – 2017-02-14
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    @A---B At 6th line, "LHS of this inequalty" may be positive or negative. If yu draw triangle of o,x,z, since y is internally divide point of x-z, you know it's wrong fact(or true when both of x,z are on first quadrant, though). – 2017-02-15

1 Answers 1

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Observe that $z-y=t(z-x)$ and $y-x=(1-t)(z-x)$.

This makes the proof much shorter.

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    That is not the point. This solution is already given in the book, I am not interested in knowing alternate solution. I want to know why my solution is incorrect. – 2017-02-13