Prove that $\frac{1}{r_{1}+1} + \frac{1}{r_{2}+1} + ... + \frac{1}{r_{n}+1} \ge \frac{n}{\sqrt[n]{r_{1}r_{2}...r_{n}}+1}$ .

Question

Let $r_{1}$,$r_{2}$, ... , $r_{n}$ be real numbers greater than or equal to $1$ . Prove that $$\frac{1}{r_{1}+1} + \frac{1}{r_{2}+1} + ... + \frac{1}{r_{n}+1} \ge \frac{n}{\sqrt[n]{r_{1}r_{2}...r_{n}}+1}$$

My Attempt:

By A.M. - H.M., $$\frac{1}{n}(\frac{1}{r_{1}+1} + \frac{1}{r_{2}+1} + ... + \frac{1}{r_{n}+1}) \ge \frac{n}{r_{1}+r_{2}+...+r_{n}+n}$$ i.e., $$\frac{1}{r_{1}+1} + \frac{1}{r_{2}+1} + ... + \frac{1}{r_{n}+1} \ge \frac{n^2}{r_{1}+r_{2}+...+r_{n}+n}$$

The inequality to prove was $\frac{1}{r_{1}+1} + \frac{1}{r_{2}+1} + ... + \frac{1}{r_{n}+1} \ge \frac{n}{\sqrt[n]{r_{1}r_{2}...r_{n}}+1}$ and now it becomes $\frac{n^2}{r_{1}+r_{2}+...+r_{n}+n} \ge \frac{n}{\sqrt[n]{r_{1}r_{2}...r_{n}}+1}$ i.e., $\frac{n}{\sqrt[n]{r_{1}r_{2}...r_{n}}+1} \ge \frac{1}{\sqrt[n]{r_{1}r_{2}...r_{n}}+1}$

Which on flipping becomes, $\frac{r_{1}+r_{2}+...+r_{n}}{n} \le \sqrt[n]{r_{1}r_{2}...r_{n}}$ which contradicts the fundamental A.M.-G.M. inequality.

What mistakes had I made during the approach and what should I be doing instead ?

Answer 1

I will give just some hints. It is just a mimic of this Cauchy proof of Am-Gm inequality.

Step 1

Prove the statement for $n=2$. Here you will use $a,b\geq 1$ and possibly Am-Gm.

Step 2

Prove the statement for $n=2^k$ with induction on $k$.

Step 3

Use backward induction i.e. go from $n$ to $n-1$. Define $r_n$ with the following equation $${1\over r_n+1} = {{1\over r_1+1} +{1\over r_2+1} +...+ {1\over r_{n-1}+1}\over n-1}$$

Carefull

Check if such $r_n$ is really $\geq 1$.

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On request @MichaelRozenberg

For $n=3$ we have $${1\over r_4+1} = {{1\over r_1+1} +{1\over r_2+1} + {1\over r_3+1}\over 3}$$

So $${{1\over r_1+1} +{1\over r_2+1} + {1\over r_3+1}+{1\over r_4+1} \over 4} = {1\over r_4+1}$$ Since we know $${{1\over r_1+1} +{1\over r_2+1} + {1\over r_3+1}+{1\over r_4+1} \over 4} \geq {1\over \sqrt[4]{r_1r_2r_3r_4}+1}$$ we have
$${1\over r_4+1} \geq {1\over \sqrt[4]{r_1r_2r_3r_4}+1}$$ then $$\sqrt[4]{r_1r_2r_3r_4} \geq r_4\implies \sqrt[3]{r_1r_2r_3}\geq r_4\implies {1\over r_4+1}\geq {1\over \sqrt[3]{r_1r_2r_3} +1}$$ and we are done.

Answer 2

There is no mistake, just your first step was too much strong that you got to prove a wrong inequality.

Your way is a similar to the following wrong reasoning.

Let we need to prove that $2>1$.

We know that $2>0$.

Thus, it's enough to prove that $0>1$, which is wrong.

Your question is: "What mistakes had I made during the approach ?"

I hope, it will help.

Answer 3

Another way.

If $r_n\rightarrow+\infty$, so the inequality is obviously true.

If $r_n=1$, so by induction it's enough to prove that: $$\frac{n-1}{\sqrt[n-1]{\prod\limits_{i=1}^{n-1}r_i}+1}+\frac{1}{2}\geq\frac{n}{\sqrt[n]{\prod\limits_{i=1}^{n-1}r_i}+1}$$ or $$\frac{n-1}{x^n+1}+\frac{1}{2}\geq\frac{n}{x^{n-1}+1},$$ where $x=\left(\prod\limits_{i=1}^nr_i\right)^{\frac{1}{n(n-1)}}\geq1$, or $f(x)\geq0,$ where $$f(x)=\frac{n-1}{x^n+1}+\frac{1}{2}-\frac{n}{x^{n-1}+1}.$$ But, $$f'(x)=-\frac{n(n-1)x^{n-1}}{\left(x^n+1\right)^2}+\frac{n(n-1)x^{n-2}}{\left(x^{n-1}+1\right)^2}=\frac{n(n-1)x^{n-2}\left(\left(x^n+1\right)^2-x\left(x^{n-1}+1\right)^2\right)}{\left(x^{n-1}+1\right)^2\left(x^n+1\right)^2}=$$ $$=\frac{n(n-1)x^{n-2}\left(x^n+1-\sqrt{x}\left(x^{n-1}+1\right)\right)\left(x^n+1+\sqrt{x}\left(x^{n-1}+1\right)\right)}{\left(x^{n-1}+1\right)^2\left(x^n+1\right)^2}=$$ $$=\frac{n(n-1)x^{n-2}\left(x^{n-\frac{1}{2}}-1\right)(\sqrt{x}-1)\left(x^n+1+\sqrt{x}\left(x^{n-1}+1\right)\right)}{\left(x^{n-1}+1\right)^2\left(x^n+1\right)^2}\geq0,$$ which says $$f(x)\geq f(1)=0.$$ Id est, it's enough to prove our inequality in the case, when a minimum point $\left(r_1,r_2,...,r_{n}\right)$ of the function $$F\left(r_1,r_2,...,r_{n}\right)=\sum_{i=1}^n\frac{1}{r_i+1}-\frac{n}{\sqrt[n]{\prod\limits_{i=1}^nr_i}+1}$$ is an inside point of $[1,+\infty)^n$.

But in this point we have $$\frac{\partial F}{\partial r_i}=0,$$ which says $$-\frac{1}{\left(r_i+1\right)^2}+\frac{\sqrt[n]{\frac{\prod\limits_{j\neq i}r_j}{r_i^{n-1}}}}{\left(\sqrt[n]{\prod\limits_{j=1}^nr_j}+1\right)^2}=0$$ or $$\frac{r_i}{\left(r_i+1\right)^2}=\frac{\sqrt[n]{\prod\limits_{j=1}^nr_j}}{\left(\sqrt[n]{\prod\limits_{j=1}^nr_j}+1\right)^2},$$ which gives that for any $i\neq j$ we have: $$\frac{r_i}{\left(r_i+1\right)^2}=\frac{r_j}{\left(r_j+1\right)^2}$$ or $$(r_i-r_j)(r_ir_j-1)=0.$$ The case $r_ir_j=1$ gives $r_i=r_j=1$ and we proved our inequality,

when even one of them is equal to $1$.

But in the case $r_i=r_j$ for any $i$ and $j$ we obtain $r_1=r_2=...=r_n,$ which gives an equality.