Why is log$(\frac{G}{a})+\frac{a}{G} \geq 1$

Asked Apr 29 '23 at 18:42

Active Apr 29 '23 at 22:42

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I am currently trying to understand Alzers proof of the arithmetic mean geometric-mean inequality. (Link)

It states that for $a\displaystyle_i, p_i \in \mathbb{R} \; \text{with} \sum p_i =1\text{ and }\mathbf{G}:= \prod a_i^{p_i}$ there exists an $k \;$ such that $a_k \leq \mathbf{G} \leq a_{k+1}$.
With this information it is now said, that $\displaystyle\int_{a_i}^{\mathbf{G}} \left(\frac{1}{t}-\frac{1}{\mathbf{G}}\right) \mathrm{d}t \geq 0 \; \; \forall i \leq k$.

When integrating I get

$$\log\left(\dfrac{G}{a_i}\right) -1 + \dfrac{a_i}{G} \geq 0$$

I ofc. know, that $\log\left(\dfrac{\mathbf{G}}{a_i}\right)\geq0$.

The proof seems so obvious but I can't figure it out and it drives me crazy.

Thanks in advance

edited Apr 29 '23 at 22:42

Math Attack

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asked Apr 29 '23 at 18:42

ruffler

2

With $x=a/G$ this is equivalent to $-\log(x) + x \ge 1$, or $\log(x) \le x-1$, which is a well-known inequality, see for example https://math.stackexchange.com/q/3188910/42969. – Martin R Apr 29 '23 at 18:57
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Does this answer your question? prove $\log(1+x)<x$ for $x>0$ – Martin R Apr 29 '23 at 19:10
Thank you very much! That was what I was looking for – ruffler Apr 30 '23 at 09:48

Why is log$(\frac{G}{a})+\frac{a}{G} \geq 1$

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