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Let $a_{1},a_{2},\cdots,a_{n}(n\ge 2)$ be postive real numbers,such that $$a_{1}a_{2}\cdots a_{n}=1$$show that $$\sum_{i=1}^{n}\left(\dfrac{1}{1+a_{i}}\right)^n\ge \dfrac{n}{2^n}$$

In fact,the function $$f(x)=\dfrac{1}{(e^x+1)^n}$$can't convex

such as $n=2$ $$f(x)=\dfrac{1}{(e^x+1)^2}\Longrightarrow f''(x)=\dfrac{2e^x(2e^x-1)}{(e^x+1)^4}$$ so Jenson inequality can't works

math110
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1 Answers1

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For clarity, here is the (invalid!) argument from convexity. Let $x_i=\log a_i$. If the function $f(x)=(1+e^x)^{-n}$ were convex for all $x$, then $\sum_i x_i=0$ and Jensen's inequality would imply

$$\frac1n \sum_{i=1}^{n}\left(\dfrac{1}{1+a_{i}}\right)^n=\frac1n \sum_{i=1}^{n} f(x_i)\ge f\left(\frac1n \sum_i x_i\right)=f(0)=\frac{1}{2^n}$$ which is the statement to be proven. But, as was noted in the OP, $f(x)$ is not everywhere convex; more precisely, it is convex only for $x\geq -\log n$. Hence Jensen's inequality is only applicable when all $x_i\geq -\log n$ i.e. none of the $a_i$ are smaller than $1/n$, which is too strong a demand for the problem's hypotheses.

Semiclassical
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  • FYI: That the argument above is invalid reflects the fact that I had initially hoped to use it as a jumping-off point for a valid convexity argument using a convex minorant (see Steele's book on Cauchy-Schwartz). This hasn't worked out yet, but I thought the above might still be a useful reference point. – Semiclassical Jun 12 '16 at 18:52
  • Cauchy-Schwarz (minus t)? – WimC Jun 12 '16 at 18:56
  • Correct. Here's the Google books link. The relevant material starts on page 96, for reference. @WimC – Semiclassical Jun 12 '16 at 19:02
  • Thank you - I was strongly misguided! – Thomas E Jun 12 '16 at 23:51
  • Since this answer doesn't actually answer the question, I'm marking it as CW since it really doesn't qualify for the bounty. – Semiclassical Jun 13 '16 at 05:09