Prove for $f(x)=x+\frac{1}{x}$

Question

Let $f(x)=x+\frac1{x}$ prove that

$|f(x)-f(1)| \le (1+\frac1{x})|x-1|$ for $x>0$

I've tried to get it from the left hand side

$|f(x)-f(1)|=|x+ \frac1{x}-2|=|\frac{x^2-2x+1}{x}|=|\frac{x-1}{x}||x-1|$

However, I can't see how to proceed from here to get $(1+\frac1{x})$ on the right hand side as well as to make that inequality appear.

Try $$|f(x)-f(1)|=|x-1+1/x-1|\leq|x-1|+|1/x-1|$$, it's the triangle inequality. — L. Milla, Jan 18 '24 at 17:02
The absolute-value signs seem to be annoying, have you tried squaring? — SchellerSchatten, Jan 18 '24 at 17:11
@CyclotomicField $|1 - \frac 1x | \leq 1 + \frac1x$ is probably what you had in mind. — mathcounterexamples.net, Jan 18 '24 at 17:14
Well we need to compare $|\frac {x-1}x|$ so $x-1$. $|\frac {x-1}x| = |1 -\frac 1x|$. If $\frac 1x < 1$ we have $|1-\frac 1x|=1-\frac 1x < 1+\frac 1x$ when $0< \frac 1x$. So this holds if $0 < \frac 1x < 1$ or in other words when $x > 1$. We can do something similar for $0 < x < 1$ and for $x=1$ the result is obvious. — fleablood, Jan 18 '24 at 17:38

score 1 · Accepted Answer · answered Jan 18 '24 at 17:15

1

Use triangle inequality: $|\frac{x-1}{x}||x-1|=|1-\frac{1}{x}||x-1| \le (1+|\frac{1}{x}|)|x-1|=[x>0]=(1+\frac{1}{x})|x-1|$

answered Jan 18 '24 at 17:15

Vasili

So it's possible to use the triangle inequality in $|1-\frac1{x}|$ while having that $|x-1|$ multiplying it ? – Ruan Carlos Jan 18 '24 at 17:37
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@RuanCarlos: yes, because if you multiply by a bigger factor, the whole expression is bigger. In other words, if $a>b>0, c>0 \implies ac > bc$ – Vasili Jan 18 '24 at 18:07
But in this case it's possible to have $|x-1|=0$ right, since x>0? – Ruan Carlos Jan 18 '24 at 18:21
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@RuanCarlos: right, that's why we have less or equal. – Vasili Jan 18 '24 at 18:22
Indeed, thank you so much! – Ruan Carlos Jan 18 '24 at 18:24

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