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I want to show that the natural logarithm is concave by only using the definition of concavity (i.e. no 2nd derivative rule or AGM inequality or other lemmas).

So I want to show that $\ln(\lambda x_1 + (1-\lambda)x_2) \geq \lambda \ln(x_1) + (1-\lambda)\ln(x_2)$ with $\lambda \in [0,1].$ So far I only could do a few steps that don't seem expedient:

$\lambda \ln(x_1) + (1-\lambda)\ln(x_2) = \ln\left(x_1^{\large\lambda} \cdot x_2^{\large 1-\lambda}\right)$ and the next thing I thought about was to show that

$h(x, y) = \lambda x + (1-\lambda)y - x^{\lambda }y^{1-\lambda} > 0 \space \space \forall x, y \in \mathbb{R}^+$ which holds and would imply the inequality (since the natural logarithm is strictly increasing) but this seems a bit cumbersome. Is there a more elegant way (only using the definition)?

Jacob
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  • Note that $h(x,x)=0$, so $h(x,y)\ge0$ is the most one can hope for. Since logarithms don't play nicely with sums in their arguments, I suggest trying to show that $e^x$ is convex (from the definition) and then conclude the desired concavity of $\ln x$ from the relationship between the graph of a function and the graph of its inverse. – Greg Martin May 04 '21 at 20:35
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    If we keep going in your direction, you're trying to show that $\lambda x_1+(1-\lambda)x_2 \ge x_1^\lambda x_2^{1-\lambda}$. This is exactly the two-variable weighted AM-GM inequality, so I'm not entirely sure what it would mean to prove it without AM-GM; whatever proof you came up with would probably be a proof of AM-GM! – Micah May 04 '21 at 20:38
  • @Micah I didn't show that yet, the reason for my question is that I want to prove the AM-GM inequality by using Jensen's inequality applied on $-\ln(x)$ which explains why I can't use the AM-GM to show it. – Jacob May 04 '21 at 20:41
  • I see, that makes sense. I think any such proof has a good chance of being circular or near-circular. Perhaps Greg's approach is the best way around this, though I haven't thought through the details precisely. I will note that while Wikipedia claims that AM-GM can be proved in this manner, it doesn't cite any sources... – Micah May 04 '21 at 21:10
  • @Jacob You can certainly prove the unweighted AM-GM inequality $\frac{a+b}{2} \ge \sqrt{ab}$ without using convexity arguments. Then, assuming you know or otherwise proved that $\ln(x)$ is continuous, use that Midpoint-Convex and Continuous Implies Convex. – dxiv Jun 01 '21 at 03:57

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One way to prove the generalized AM-GM inequality $$\lambda x+(1-\lambda)y\ge x^\lambda y^{1-\lambda}\tag1$$ for all $x$, $y$ positive and $\lambda\in[0,1]$ is to note that the inequality is homogeneous in $x$, $y$ (meaning the same inequality obtains if we replace $x$, $y$ with $cx$, $cy$); hence we can impose the additional assumption $$x^\lambda y^{1-\lambda}=1.\tag2$$ Using the inequality $\log t\le t-1$ the LHS of (1) is $$\lambda x+(1-\lambda)y\ge\lambda(\log x+1) + (1-\lambda)(\log y+1)=[\lambda\log x+(1-\lambda )\log y]+1.\tag3$$ By assumption (2), the bracketed part of the RHS of (3) equals zero, which completes the proof.

grand_chat
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