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let $x_{i}>0(i=1,2,\cdots,n)$,prove or disprove $$\dfrac{n-1+x_{2}x_{3}\cdots x_{n}}{1+(n-1)x_{1}}+\dfrac{n-1+x_{1}x_{3}\cdots x_{n}}{1+(n-1)x_{2}}+\cdots+\dfrac{n-1+x_{1}x_{2}\cdots x_{n-1}}{1+(n-1)x_{n}}\ge n$$

I try to use Cauchy-Schwarz inequality $$LHS\cdot\left(n+(n-1)\sum_{i=1}^{n}x_i\right)\ge \left(\sum_{cyc}\sqrt{n-1+x_{2}x_{3}\cdots x_{n}}\right)^2$$ we need show $$\left(\sum_{cyc}\sqrt{n-1+x_{2}x_{3}\cdots x_{n}}\right)^2\ge n^2+n(n-1)\sum_{i=1}^{n}x_{i}$$ it seem this last inequality not right,because from Maclaurin's inequality it's Reverse inequality

math110
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1 Answers1

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Partial answer that perhaps someone can complete it :

First define :

$$f\left(x\right)=\frac{\left(\left(n-1\right)x+a\right)}{x+\left(n-1\right)x^{2}}$$

For $a\geq 0.25$ and $n\geq 3$ we have for $x\in(-\infty,\infty)$ :

$$h\left(x\right)=\frac{d}{dx}\left(\frac{d}{dx}\left(f\left(e^{x}\right)\right)\right)>0$$

For details on the second derivatives see Wolfram alpha .

Now conclude using Jensen's inequality as the function $h(x)$ is convex . The rest is smooth because it's a simple one variable inequality .

In fact we can completes it partially using RCF corollary see https://link.springer.com/content/pdf/10.1186/1029-242X-2011-101.pdf .

For that we can use a little $\varepsilon$ such that $x_1x_2\cdots x_n\geq \varepsilon>0$

To conclude definitively we can use Jensen's inequality on $f(x)$:

$$f''(x)>0$$

For $0<a=x_1x_2\cdots x_n\leq \varepsilon $ and $n\geq 3 $ .

I can add details if necessary .

Miss and Mister cassoulet char
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  • So, keeping $a$ fixed, $f_a(e^t)$ is convex in $t$. Hence, $\sum f_a(t_i)$ is convex in $(t_1,\ldots,t_n)$. The convexity then applies to the linear subspace $\sum t_i=\ln a$, which yields $\sum f_a(e^{t_i})\le n f_a(e^t)$ where $nt = \sum t_i = \ln a$. Then, all that remains is showing $f_a(x)\ge 1$ for $a=x^n$. – Einar Rødland May 01 '22 at 07:30