Let $\{a_1,a_2,\cdots,a_n\}\subset[0,1]$ $(n\geq 2)$ . Prove that $$\frac{1}{1+a_1+a_2}+\frac{1}{1+a_2+a_3}+\cdots+\frac{1}{1+a_{n-1}+a_n}+\frac{1}{1+a_n+a_1}\leq\frac{n}{1+2\sqrt[n]{a_1a_2\cdots a_n}}$$ sqing wrote: Let $a,b,c,d \in [0,1] .$ Prove that$$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+d}+\frac{1}{1+d+a}\leq \frac{4}{1+2\sqrt[4]{abcd}}$$

sqing wrote: Let $a,b,c,d \in [0,1] .$ Prove that$$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+d}+\frac{1}{1+d+a}\leq \frac{4}{1+2\sqrt[4]{abcd}}$$ Just a little calculation, not a hard problem.

sqing wrote: Let $\{a_1,a_2,\cdots,a_n\}\subset[0,1]$ $(n\geq 2)$ . Prove that $$\frac{1}{1+a_1+a_2}+\frac{1}{1+a_2+a_3}+\cdots+\frac{1}{1+a_{n-1}+a_n}+\frac{1}{1+a_n+a_1}\leq\frac{n}{1+2\sqrt[n]{a_1a_2\cdots a_n}}$$ @sqing, can you post the solution for this problem? Very nice one, thanks!

EthanWYX2009 wrote: $$\frac{4}{1+2\sqrt[4]{abcd}}-(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+d}+\frac{1}{1+d+a})$$$$=\int_{0}^1(4x^{2\sqrt[4]{abcd}}-x^{a+b}-x^{b+c}-x^{c+d}-x^{d+a})dx$$$$=\int_{0}^1(4x^{2\sqrt[4]{abcd}}-(x^a+x^c)(x^b+x^d))dx$$$$\ge\int_{0}^1\left(4x^{\sqrt {ac}+\sqrt {bd}}-(2x^{\sqrt {ac}})\cdot(2x^{\sqrt {bd}})\right)dx=0.\blacksquare$$ Can you explain, why does $x^a+x^c\leq2x^{\sqrt{ac}}$?

arqady wrote: EthanWYX2009 wrote: $$\frac{4}{1+2\sqrt[4]{abcd}}-(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+d}+\frac{1}{1+d+a})$$$$=\int_{0}^1(4x^{2\sqrt[4]{abcd}}-x^{a+b}-x^{b+c}-x^{c+d}-x^{d+a})dx$$$$=\int_{0}^1(4x^{2\sqrt[4]{abcd}}-(x^a+x^c)(x^b+x^d))dx$$$$\ge\int_{0}^1\left(4x^{\sqrt {ac}+\sqrt {bd}}-(2x^{\sqrt {ac}})\cdot(2x^{\sqrt {bd}})\right)dx=0.\blacksquare$$ Can you explain, why does $x^a+x^c\leq2x^{\sqrt{ac}}$? I'm sorry, I think this is a wrong prove(I saw this in SZM yesterday but it was deleted now). Can anyone fix this prove?

@sqing,can you post the answer to the generalized problem? Thanks a lot

sqing wrote: Let $\{a_1,a_2,\cdots,a_n\}\subset[0,1]$ $(n\geq 2)$ . Prove that $$\frac{1}{1+a_1+a_2}+\frac{1}{1+a_2+a_3}+\cdots+\frac{1}{1+a_{n-1}+a_n}+\frac{1}{1+a_n+a_1}\leq\frac{n}{1+2\sqrt[n]{a_1a_2\cdots a_n}}$$ Sorry but this is wrong . Try $n=7$ with $1,1,0.9,0.8,0.7,0.75,0.9$.

Try this code to find examples. Try several times if the result is still positive.

RainbowNeos wrote: sqing wrote: Let $\{a_1,a_2,\cdots,a_n\}\subset[0,1]$ $(n\geq 2)$ . Prove that $$\frac{1}{1+a_1+a_2}+\frac{1}{1+a_2+a_3}+\cdots+\frac{1}{1+a_{n-1}+a_n}+\frac{1}{1+a_n+a_1}\leq\frac{n}{1+2\sqrt[n]{a_1a_2\cdots a_n}}$$ Sorry but this is wrong . Try $n=7$ with $1,1,0.9,0.8,0.7,0.75,0.9$. Actually I want it to be right, then it will be a really really nice problem. Can we prove it when ${n}$ is even?

EthanWYX2009 wrote: RainbowNeos wrote: sqing wrote: Let $\{a_1,a_2,\cdots,a_n\}\subset[0,1]$ $(n\geq 2)$ . Prove that $$\frac{1}{1+a_1+a_2}+\frac{1}{1+a_2+a_3}+\cdots+\frac{1}{1+a_{n-1}+a_n}+\frac{1}{1+a_n+a_1}\leq\frac{n}{1+2\sqrt[n]{a_1a_2\cdots a_n}}$$ Sorry but this is wrong . Try $n=7$ with $1,1,0.9,0.8,0.7,0.75,0.9$. Actually I want it to be right, then it will be a really really nice problem. Can we prove it when ${n}$ is even? I tried $n=8,10,12,14$ (and $9,11,13$) but it is still wrong. The following examples are found by Python. $n=8$: $0.62671984, 0.72346595, 0.89023479, 1. , 1. , 0.89024881,0.72347815, 0.62672308$ $n=9$: $1. , 0.83431808, 0.65122964, 0.5574767, 0.55746876, 0.65123224, 0.83431696, 1., 1., $ $n=10$: $1., 1., 1., 1., 0.7920143, 0.60306064, 0.51409886, 0.51409615, 0.60305219, 0.79201053$ $n=11$: $0.62264728, 0.50688512, 0.47239244, 0.50691381, 0.62269067, 0.82581801, 1. , 1., 1., 1., 0.82578819$ $n=12$: $1., 1., 1., 1., 1., 1., 0.7330254 , 0.54203096, 0.4614578, 0.46144872, 0.54204714, 0.73305851$ $n=13$: $1., 0.83603591, 0.62336793, 0.49354267, 0.43987623, 0.43987264, 0.49352674, 0.62334548, 0.83603007, 1., 1., 1., 1. $ $n=14$: $0.42509665, 0.42512584, 0.47323456, 0.59309378, 0.80366905, 1., 1., 1., 1., 1., 1., 0.80356986, 0.59297313,0.47315825$ So I guess that this inequality is only true for $n\leq 6$.

Clearly, the inequality holds when any of $a,b,c,d$ are $0$. So now set $a,b,c,d\in(0,1]$. Note that the function $f(x)=\frac1{k+e^x}$ is concave for $k\ge1,x\le0$ since $f''(x)=\frac{e^x(e^x-k^2)}{\left(k+e^x\right)^4}\le0$. Then it follows from Jensen's Inequality that $$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+d}+\frac{1}{1+d+a}$$$$\leq\frac2{1+b+\sqrt{ac}}+\frac2{1+d+\sqrt{ac}}$$$$\le\frac4{1+\sqrt{ac}+\sqrt{bd}}$$which is at most $\frac4{1+2\sqrt[4]{abcd}}$ by AM-GM.

It's really a “miracle” to see the four variable inequality in 2023......