Consider $x,y$ and $z$ in $\mathbb R$. Suppose $|x-z| \geq 1$. Show that $|x-y|e^{-|x-y|^2} + |z-y|e^{-|z-y|^2} \leq 2 |y| e^{-|y|^2} $.
I tried mean value thm but no luck. I appreciate some hints or help.
Show that $|x-y|e^{-|x-y|^2} +|z-y|e^{-|z-y|^2} \leq 2 |y| e^{-|y|^2} $
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real-analysis
analysis
inequality
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0Well, $|x-y|^{2}=(x-y)^{2}$, so you could try to differentiate the LHS to maximise it? – 2017-02-17
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0@DanielLittlewood Thanks. I also did it but it got more complicated. – 2017-02-17
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0Isn't that wrong for $y=0$ ? – 2017-02-17
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0@MartinR Indeed, $(1/2,0,-1/2)$ appears to be a counterexample – 2017-02-17
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0@DanielLittlewood Sorry, the original question should have absolute value. Just edited. – 2017-02-17
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0@pythh: That is still wrong if $y=0$ and $x, z \ne 0$. – 2017-02-17