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Let $x_1,x_2,\cdots x_n,x_{n+1}$ be any real numbers greater than or equal to $1$.

Then for $n\ge 2,$ I was trying to verify the validity of the inequality $$\frac{n-1}{n}\sum_{k=1}^n\frac{1}{1+x_k}+\frac{1}{1+x_{n+1}}+\frac{1}{1+x_1.x_2.\cdots.x_n}\ge \frac{n}{n+1}\sum_{k=1}^{n+1}\frac{1}{1+x_k}+\frac{1}{1+x_1x_2\cdots x_nx_{n+1}}.$$ May I kindly seek your suggestions?

Same result can be tried when $x_1,x_2,\cdots x_n,x_{n+1}$ are non-negative real numbers less than or equal to $1$.

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An attempt towards the solution: We need to prove $$(n^2-1)\sum_{k=1}^n\frac{1}{1+x_k}+\frac{n(n+1)}{1+x_{n+1}}+\frac{n(n+1)}{1+x_1.x_2.\cdots.x_n}\ge n^2\sum_{k=1}^{n+1}\frac{1}{1+x_k}+\frac{n(n+1)}{1+x_1x_2\cdots x_nx_{n+1}},$$ which is équivalent to $$\frac{n}{1+x_{n+1}}+\frac{n(n+1)}{1+x_1.x_2.\cdots.x_n}-\frac{n(n+1)}{1+x_1x_2\cdots x_nx_{n+1}}-\sum_{k=1}^n\frac{1}{1+x_k}\geq 0.$$ When $x_{n+1}=1$ the result is true. But for very large values of $x_n,x_{n+1} (x_{n}\rightarrow\infty, x_{n+1}\rightarrow\infty)$ the above inequality is not true. Hope my argument is correct!

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    $\begingroup$ Right, so for example, using $n=2$, the inequality fails for $(x_1,x_2,x_3)=(1,12,12)$, and it also fails for $(x_1,x_2,x_3)=(1,\frac{1}{2},\frac{1}{2})$. $\endgroup$ – quasi Sep 23 '18 at 13:39
  • $\begingroup$ Thank you. I was wondering if the coefficients of sum in RHS and LHS can be redefined to make the inequality valid? $\endgroup$ – user159888 Sep 24 '18 at 4:27
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    $\begingroup$ What is the intended application? $\endgroup$ – quasi Sep 24 '18 at 5:43
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The inequality is equal to

$$\frac{n-1}{n}\Big(\sum_{k=1}^n\frac{1}{1+x_k}\Big)+\frac{1}{1+x_{n+1}}+\frac{1}{1+x_1.x_2.\cdots.x_n} \ge \frac{n}{n+1}\Big(\sum_{k=1}^{n}\frac{1}{1+x_k}\Big)+\frac{n}{n+1}\Big(\frac{1}{1+x_{n+1}}\Big)+\frac{1}{1+x_1x_2\cdots x_nx_{n+1}}$$


We know that $\frac{n-1}{n}>\frac{n}{n+1}$ hence

$$\frac{n-1}{n}\Big(\sum_{k=1}^n\frac{1}{1+x_k}\Big)\ge\frac{n}{n+1}\Big(\sum_{k=1}^{n}\frac{1}{1+x_k}\Big)$$

Since $1>\frac{n}{n+1}$ $$\frac{1}{1+x_{n+1}}\ge\frac{n}{n+1}\Big(\frac{1}{1+x_{n+1}}\Big)$$

Since $a\le b \implies \frac{1}{a} \ge \frac{1}{b}$ hence

$$\frac{1}{1+x_{n+1}}\ge\frac{1}{1+x_1.x_2.\cdots.x_n.x_{n+1}}$$

Adding the three equations above gives us the inequality we seek. $\ _\square$

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  • $\begingroup$ The assumption $\frac{n-1}{n}>\frac{n}{n+1}$ is incorrect in the above answer. Cross multiply and check! $\endgroup$ – user159888 Sep 23 '18 at 7:02
  • $\begingroup$ @user159888 I see it sorry $\endgroup$ – Vee Hua Zhi Sep 23 '18 at 7:06

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