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 $1$.


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!

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ What is the intended application? $\endgroup$ – quasi Sep 24 '18 at 5:43

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


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


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.