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Standard Cauchy induction works just fine. We first prove it for two numbers: It takes the form $a,b\ge 1\Rightarrow \frac 1{a^2+1}+\frac 1{b^2+1}\ge \frac 2{ab+1}$. After expanding everything, this reduces to $2ab(ab-1)\le (a^2+b^2)(ab-1)$, which is true by AM-GM $(*)$. The next step is proving it for $n=2^k$. This is done easily by induction: we add two inequalities with $2^{k-1}$ variables each, and then apply the inequality for two variables to get the result $(**)$. After this, we prove that $P(n)\Rightarrow P(n-1)$. Assuming the inequality is true for $n$ variables, take $r_n=\sqrt[n-1]{r_1r_2\ldots r_{n-1}}$. We get $\sum_{i=1}^{n-1}\frac 1{r_i+1}+\frac 1{\sqrt[n-1]{r_1\ldots r_{n-1}}+1}\ge \frac n{\sqrt[n-1]{r_1\ldots r_{n-1}}+1}$. This proves the inequality for $n-1$ variables $(***)$. $(*),(**),(***)$ complete the induction.

Jensen inequality works even easier.

Quote: Jensen inequality works even easier How? would you send your solution I couldn't solve it with jensen

Yes . Jensen for the function $f(x)=\frac{1}{x+1}$ doesnot work.

Try $f(x) = \frac{1}{e^x + 1}$.

I have just seen this theorem and I think I can prove the inequality by showing that: $\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}\geq \frac{1}{\sqrt{r_{1}r_{2}}+1}+\frac{1}{\sqrt{r_{1}r_{2}}+1}$ $(r_{1}+r_{2}+2)(\sqrt{r_{1}r_{2}}+1)\geq 2(r_{1}+1)(r_{2}+1)$ $(r_{1}+r_{2}-2\sqrt{r_{1}r_{2}})\sqrt{r_{1}r_{2}}\geq (r_{1}+r_{2}-2\sqrt{r_{1}r_{2}})$ $\sqrt{r_{1}r_{2}}\geq 1$ (True by hypotesis) Changing the minimum and the maximum element with their GM $\infty$ times all the elements become equal, so I have: $\sum_{i=1}^{n}\frac{1}{r_{i}+1}\geq \sum_{i=1}^{n}\frac{1}{\sqrt[n]{x_{1}...x_{n}}+1}=\frac{n}{\sqrt[n]{x_{1}...x_{n}}+1}$

Jensen:

Yeah....grobber's solution is just like the A-M G-M induction proof...Great job!!.....Anyway the proof using jensen's inequality is an elegant one and easier to realise.

@sayantanchakraborty basically that's how one can prove Jensen's inequality.

Smoothing also works, note that when $x_1x_2>1$ is fixed, the function $\frac {1}{x_1+1}+\frac {1}{x_2+1}=\frac {x_1+x_2+2}{x_1x_2+x_1+x_2+1}$ decreases as $x_1+x_2$ decrease. This happens when we apply the replacement for $r_i<r'<r_j$($r'=\sqrt[n]{r_{1}r_{2}\cdots r_{n}}$) as $(r_i,r_j)\rightarrow (r',\frac {r_ir_j}{r'})$.

Regular induction also works; it's so easy it feels like cheating. The $n=2$ case is equivalent to $(r_1r_2-1)(r_1-r_2)^2\ge 0$ which is obvious and has equality $r_1=r_2$. For $n$ numbers, assuming the $n-1$ case is correct, order them such that $r_n$ is the least, and let $k=r_1r_2\dots r_{n-1}$. Now by the $n-1$ case on all of them except $r_n$ we would like to show that $\tfrac{n-1}{\sqrt[n-1]{k}+1}+\tfrac{1}{r_n+1}\ge \tfrac{n}{\sqrt[n]{kr_n}+1}$. This follows since $\tfrac{n-1}{\sqrt[n-1]{k}+1}\ge \tfrac{n-1}{\sqrt[n]{kr_n}+1}$ and $\tfrac{1}{r_n+1}\ge \tfrac{1}{\sqrt[n]{kr_n}+1}$ both reduce to $k\ge r_n^{n-1}$, which is true.

Sorry to revive . I think that there is a mistake here ! Please check it. This follows since $\tfrac{n-1}{\sqrt[n-1]{k}+1}\ge \tfrac{n-1}{\sqrt[n]{kr_n}+1}$ and $\tfrac{1}{r_n+1}\ge \tfrac{1}{\sqrt[n]{kr_n}+1}$ both reduce to $k\ge r_n^{n-1}$, which is true. [/quote]

orl wrote: Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. Prove that \[ \frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n}+1} \geq \frac{n}{ \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \] Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. For $\lambda\ge 1$, prove that \[ \frac{1}{\lambda r_{1} + 1} + \frac{1}{\lambda r_{2} + 1} + \cdots +\frac{1}{\lambda r_{n}+1} \geq \frac{n}{\lambda \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \]

sqing wrote: Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. For $\lambda\ge 1$, prove that \[ \frac{1}{\lambda r_{1} + 1} + \frac{1}{\lambda r_{2} + 1} + \cdots +\frac{1}{\lambda r_{n}+1} \geq \frac{n}{\lambda \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \] Jensen with $f(x) = \frac1{\lambda e^x + 1}$, $f''(x) = \frac{\lambda e^x(\lambda e^x-1)}{(\lambda e^x+1)^3} > 0$.

Vrangr wrote: sqing wrote: Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. For $\lambda\ge 1$, prove that \[ \frac{1}{\lambda r_{1} + 1} + \frac{1}{\lambda r_{2} + 1} + \cdots +\frac{1}{\lambda r_{n}+1} \geq \frac{n}{\lambda \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \] Jensen with $f(x) = \frac1{\lambda e^x + 1}$, $f''(x) = \frac{\lambda e^x(\lambda e^x-1)}{(\lambda e^x+1)^3} > 0$. Thanks.

sqing wrote: Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. For $\lambda\ge 1$, prove that \[ \frac{1}{\lambda r_{1} + 1} + \frac{1}{\lambda r_{2} + 1} + \cdots +\frac{1}{\lambda r_{n}+1} \geq \frac{n}{\lambda \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \] Take in the initial inequality $r_i=\lambda x_i$ where $\lambda\ge 1\wedge x_i\ge 1$ for all $i\in\lbrace 1,2,...,n\rbrace$. Actually weaker assumptions: $\lambda>0\wedge \forall_{i\in\lbrace 1,2,...,n\rbrace}\ x_i\ge \frac{1}{\lambda}$ are sufficient. Then you get \[ \frac{1}{\lambda x_{1} + 1} + \frac{1}{\lambda x_{2} + 1} + \cdots +\frac{1}{\lambda x_{n}+1} \geq \frac{n}{\lambda \sqrt[n]{x_{1}x_{2} \cdots x_{n}}+1}. \]

For storage (same smoothing proof as others have pointed out). 1998 A2 wrote: Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. Prove that \[ \frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n}+1} \geq \frac{n}{ \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \] Cute We prove the following interim step. Lemma. Let $a, b \geqslant 1$ be real numbers. Then $$\frac{1}{a^2+1}+\frac{1}{b^2+1} \geqslant \frac{2}{ab+1}.$$ Proof. Clearing denominators, we need to show that $(2+a^2+b^2)(1+ab)-2(1+a^2)(1+b^2) \geqslant 0$. Notice that $(1+a^2)(1+b^2)=(1+ab)^2+(a-b)^2$, so the previous expression factors as $(ab-1)(a-b)^2$ and the claim follows. $\blacksquare$ Now we prove the result by induction on $n \geqslant 2$, with base case following from the earlier lemma. Assume the result holds upto all integer less than $n$. For $(r_1, \dots, r_n) \in \mathbb{R}^n$ with $r_i \ge 1$, define $f_n(r_1, \dots, r_n)=\left(\sqrt[n-1]{r_2\dots r_n}, \dots, \sqrt[n-1]{r_1 \dots r_{n-1}}\right)$ and $g(r_1, \dots, r_n)=\sum^n_{i=1} \frac{1}{r_i+1}$. Claim. For any such $n$-tuple, we have $g(r_1, \dots, r_n) \geqslant g(f_n(r_1, \dots, r_n))$. Proof. By induction hypothesis, we have the $n$ inequalities for $j=1, \dots, n$: $$\sum^n_{i=1, i \ne j} \frac{1}{r_i+1} \ge \frac{(n-1)}{\sqrt[n-1]{r_1\dots r_{j-1}r_{j+1}\dots r_n}+1}$$and adding them all proves the claim. $\blacksquare$ Thus, for any $k \in \mathbb{N}$ we have $g(r_1, \dots, r_n) \geqslant g(f_n^k(r_1, \dots, r_n))$ (here $f^k$ is the composition $k$ times). Taking limits as $k \rightarrow \infty$, since $\frac{\max(r_1, \dots, r_n)}{\min(r_1, \dots, r_n)}$ converges to $1$, we see that the RHS converges to $g(r, \dots , r)$ where $r^n=r_1\dots r_n$. Thus, the desired inequality follows. $\blacksquare$

It's easy to see that the inequality works for $r_1 = r_2 = ... = r_n$. Let $C$ is $\sqrt[n]{r_{1}r_{2}...r_{n}}$. Suppose that not all $r_i$ are equal $C$. We will make operations to increase the number of $r_i = C$. So, we can take $r_i < C$ and $r_j > C$. Let's exchange $r_i$ for $kr_i$ and $r_j$ for $\frac{r_j}{k}$ where $k$ is $\frac{C}{r_i}>1$. To prove this problem we only need to prove this inequality: $$ \frac{1}{r_1 + 1} + \frac{1}{r_2 + 1} \geq \frac{1}{kr_1} + \frac{1}{\frac{r_2}{k} + 1}$$$$ \frac{r_1+r_2+2}{(r_1+1)(r_2+1)} \geq \frac{kr_1+\frac{r_1}{k}+2}{(kr_1+1)(\frac{r_2}{k}+1)}$$$$(r_1 + r_2 + 1)(r_1r_2 + kr_1 + \frac{r_2}{k} + 1) \geq (kr_1 + \frac{r_2}{k} + 2)(r_1r_2 + r_1 + r_2 + 1)$$$$r_1^2r_2 + r_1r_2^2 + kr_1 + \frac{r_2}{k} \geq kr_1^2r_2 + \frac{r_1r_2^2}{k} + r_1 + r_2$$$$(r_1r_2 - 1)(k - 1)(\frac{r_2}{k} - r_1) \geq 0$$So, we can get $r_1 = r_2 = ... = r_n = C$, Q.E.D.

We will prove this by induction on $n$. First, we prove the base case $n=2$. Let $r_1=x_1^2, r_2=x_2^2$. Then the inequality we must prove is $$\frac{1}{x_1^2+1}+\frac{1}{x_2^2+1}\ge \frac{2}{x_1x_2+1}.$$Multiplying everything by $(x_1^2+1)(x_2^2+1)(x_1x_2+1)$ gives us we must prove $$(x_1x_2+1)((x_1^2+1)+(x_2^2+1))\ge 2(x_1^2+1)(x_2^2+1).$$Subtracting the RHS from the LHS and factoring gives $$(x_1x_2-1)(x_1-x_2)^2$$which is obviously always nonnegative. Now we use smoothing. Claim: Let $\epsilon$ be any real number greater than or equal to $1$, and let $x,y$ be greater than or equal to $1$ such that $y\ge x \epsilon^2$. Then $$\frac{1}{x+1}+\frac{1}{y+1}\ge \frac{1}{x\epsilon+1}+\frac{1}{\frac{y}{\epsilon}+1}.$$Proof. Rearranging and combining fractions, we want to prove $$\frac{1}{x+1}-\frac{1}{x\epsilon+1}\ge \frac{1}{\frac{y}{\epsilon}+1}-\frac{1}{y+1} \Rightarrow \frac{x\epsilon -x}{(x+1)(x\epsilon+1)}\ge \frac{\frac{y}{\epsilon}-y}{(\frac{y}{\epsilon}+1)(y+1)}.$$Cancelling a factor of $\epsilon-1$ on the numerator gives $$\frac{x}{(x+1)(x\epsilon+1)}\ge \frac{\frac{y}{\epsilon}}{(\frac{y}{\epsilon}+1)(y+1)}.$$However, notice that the derivative of $\frac{x}{(x+1)(x\epsilon+1)}$ is equal to $-\frac{\epsilon x^2-1}{(x+1)^2(\epsilon x+1)^2}\le 0$ (for $x\ge 1$), so the LHS is nonincreasing. As $x\le \frac{y}{\epsilon^2}$, $$\frac{x}{(x+1)(x\epsilon+1)}\ge \frac{\frac{y}{\epsilon^2}}{(\frac{y}{\epsilon^2}+1)(\frac{y}{\epsilon}+1)}$$so we simply need to prove that $$\frac{\frac{y}{\epsilon^2}}{(\frac{y}{\epsilon^2}+1)(\frac{y}{\epsilon}+1)} \ge \frac{\frac{y}{\epsilon}}{(\frac{y}{\epsilon}+1)(y+1)}\Rightarrow \frac{1}{\frac{y}{\epsilon}+\epsilon}\ge \frac{1}{y+1}\Rightarrow \frac{1}{\epsilon}(\epsilon-1)(y-\epsilon)\ge 0,$$which is clearly true. $\blacksquare$ Returning to the problem, suppose $n=k$, let $G=\sqrt[k]{r_1r_2\ldots r_k}$ and WLOG $r_1\le G\le r_2$. Then apply the claim with $x=r_1, y=r_2, \epsilon=\text{min}\left(\frac{G}{r_1}, \frac{r_2}{G}\right)$. Then notice that one of $r_1,r_2$ has become equal to $G$, and the other has been turned into $\frac{r_1r_2}{G}$ (as the product of the numbers stays the same). Thus we need to prove $$\frac{1}{G+1}+\frac{1}{\frac{r_1r_2}{G}+1}+\frac{1}{r_3+1}\ldots +\frac{1}{r_k+1}\ge \frac{k}{G+1}\Rightarrow \frac{1}{\frac{r_1r_2}{G}+1}+\frac{1}{r_3+1}\ldots +\frac{1}{r_k+1}\ge \frac{k-1}{G+1}.$$However, as this is a case of the inequality with $n=k-1$, this directly follows from the induction hypothesis, and we are done.

Here's my awful solution. Firstly WLOG $r_n\geq r_{n-1}\geq \ldots\geq r_2\geq r_1$. We use induction. We claim that for any $1\leq i\leq n$, $$\frac{1}{r_i+1}+ \frac{i-1}{ \sqrt[i-1]{r_{1}r_{2} \cdots r_{i-1}}+1}\geq \frac{i}{ \sqrt[i]{r_{1}r_{2} \cdots r_{i}}+1}.$$Let $r_i=x^i$ and $\sqrt[i-1]{r_{1}r_{2} \cdots r_{i-1}}=y^i$. Hence, we need \begin{align*} \frac{1}{x^i+1}+ \frac{i-1}{ y^i+1}&\geq \frac{i}{xy^{i-1}+1}\Longleftrightarrow \\ ((i-1)x^i+y^i+i)(xy^{i-1}+1)&\geq i(x^i+1)(y^i+1)\Longleftrightarrow \\ xy^{2i-1}+ixy^{i-1}+(i-1)x^{i+1}y^{i-1}&\geq ix^iy^i+x^i+(i-1)y^i\Longleftrightarrow \\ y^{i}+i+(i-1)x^{i}&\geq ix^{i-1}y+\frac{x^i+(i-1)y^i}{xy^{i-1}}. \end{align*}Now notice that \begin{align*}x^{i}+i+(i-1)y^{i}&\geq ixy^{i-1}+\frac{x^i+(i-1)y^i}{xy^{i-1}} \Longleftrightarrow\\ (x^{i}+(i-1)y^{i})\left(1-\frac{1}{xy^{i-1}}\right)+i(1-xy^{y-1})&\geq 0 \Longleftrightarrow\\ (xy^{y-1}-1) (\frac{x^{i}+(i-1)y^{i}}{xy^{i-1}}-i)&\geq 0, \end{align*}which holds by AM-GM and as $x,y\geq 1$. Now, I show that always \begin{align*} y^{i}+i+(i-1)x^{i}- ix^{i-1}y-\frac{x^i+(i-1)y^i}{xy^{i-1}}&\geq x^{i}+i+(i-1)y^{i}- ixy^{i-1}-\frac{x^i+(i-1)y^i}{xy^{i-1}}\Longleftrightarrow\\ y^{i}+(i-1)x^{i}- ix^{i-1}y&\geq x^{i}+(i-1)y^{i}- ixy^{i-1}\Longleftrightarrow\\ (i-2)(x^{i}-y^{i})&\geq ixy(x^{i-2}-y^{i-2}). \end{align*}If $i=1$ and $i=2$, then the inequality above is obvious. Now for $i\geq 3$, \begin{align*}(i-2)(x^{i}-y^{i})\geq ixy(x^{i-2}-y^{i-2})\Longleftrightarrow\\ (x-y)((i-2)(x^{i-1}+x^{i-2}y+\ldots+xy^{i-2}+y^{i-1})-i(x^{i-2}y+x^{i-3}y^2+\ldots+x^2y^{i-3}+xy^{i-2}))\geq 0\Longleftrightarrow\\ (x-y)((i-2)(x^{i-1}+y^{i-1})-2(x^{i-2}y+x^{i-3}y^2+\ldots+x^2y^{i-3}+xy^{i-2}))\geq 0, \end{align*}which is true by separate Muirhead's and as $x^i=r_i\geq \sqrt[i-1]{r_1r_2\ldots r_{i-1}}=y^i$. Hence, we build up our desired inequality. Note that the equality holds iff $r_1=r_2=\ldots=r_n$.

Let $a_i=\ln r_i\ge0$ and $f(x)=\frac1{1+e^x}$, then we want to prove that: $$\sum_{i=1}^nf(a_i)\ge nf\left(\frac1n\sum_{i=1}^na_i\right),$$which is trivial by Jensen since $f''(x)=\frac{e^x\left(e^x-1\right)}{\left(1+e^x\right)^3}\ge0$.

By Titu Inequality we have: LHS> n^2/(n+r1+...rn). .which is By applying AM-GM in the denominator> n^2/(n(r1...rn)^(1/n)+n) which is equivalent to: n/((r1...rn)^(1/n)+1)=RHS. Hence Proved.

Jenson inequality

Let function $f(x)=\tfrac{1}{e^x+1}$. We have $\tfrac{1}{r_1+1}=f(\ln(r_1)).$ Since $r_1\ge 1$, $\ln(r_1)$ is nonnegative. When $x\ge 0$, we have \begin{align*} f'(x) &= \frac{-e^x}{(e^x+1)^2} \\ f''(x) &= \frac{a(a-1)}{(a+1)^3} \end{align*}where $a=e^x$. Since $x\ge 0$, we have $a\ge 1$, so $f(x)$ is convex. Therefore, $f(\ln(r_1))+f(\ln(r_2))+\dots+f(\ln(r_n))\ge nf(\ln(\sqrt[n]{r_1r_2r_3\dots r_n}))$ which is the desired inequality.

The base case $n=1$ is clearly true as $LHS = RHS$. If $n=2$: $$\frac{1}{r_1 +1 } +\frac{1}{r_2 + 1} \geq \frac{2}{\sqrt{r_1 r_2} + 1}$$Let $(\sqrt{r_1}, \ \sqrt{r_2}) = (a,b)$. Then we need to show $$\frac{1}{a^2+1}+\frac{1}{b^2 + 1}\geq \frac{2}{ab + 1}$$which is equivalent to $$(b^2 + 1)(ab+1) + (a^2 + 1)(ab+1) \geq 2(a^2+1)(b^2 +1)$$$$ab^3 + 2ab + a^3b \geq 2a^2b^2 + a^2 + b^2$$$$ab(a+b)^2 \geq (a+b)^2$$$$ab \geq 1$$which is true. Suppose the statement is true for $n$. We will prove for $2n$. \begin{align*} \sum ^ {2n}_{i=1} \frac{1}{r_i + 1} &= \sum ^{n}_{i=1} \frac{1}{r_i + 1} + \sum^{2n} _{i = n+1} \frac{1}{r_i + 1}\\ &\geq \frac{n}{\sqrt[n]{r_1 r_2 \ldots r_n} + 1} + \frac{n}{\sqrt[n]{r_{n+1} r_{n+2} \ldots r_{2n}}+1}\\ &\geq \frac{2n}{\sqrt{(r_1 r_2 \ldots r_n)^\frac 1n} \cdot \sqrt{(r_{n+1} r_{n+2} \ldots r_{2n})^\frac 1n} + 1}\\ &= \frac{2n}{(r_1 r_2 \ldots r_{2n})^\frac{1}{2n} + 1}. \end{align*}We now prove for $n-1$. Let $r_n = \sqrt[n-1]{r_1 r_2 \ldots r_{n-1}}$. \begin{align*} &\sum ^{n}_{i=1} \frac{1}{r_i + 1} \geq \frac{n}{(r_1 r_2 \ldots r_n)^\frac 1n + 1}= \frac{n}{[(r_1 r_2 \ldots r_{n-1})(r_1 r_2 \ldots r_{n-1})^\frac{1}{n-1}]^\frac{1}{n} + 1} = \frac{n}{\sqrt[n-1]{r_1 r_2 \ldots r_{n-1}} + 1}\\ \implies \ & \frac{1}{r_n +1} + \sum^{n-1}_{i=1} \frac{1}{r_i+1} \geq \frac{n-1}{\sqrt[n-1]{r_1 r_2 \ldots r_{n-1}} + 1} + \frac{1}{\sqrt[n-1]{r_1 r_2 \ldots r_{n-1}} + 1} = \frac{n-1}{\sqrt[n-1]{r_1 r_2 \ldots r_{n-1}} + 1} + \frac{1}{r_n + 1}\\ \implies \ & \sum^{n-1}_{i=1} \frac{1}{r_i+1} \geq \frac{n-1}{\sqrt[n-1]{r_1 r_2 \ldots r_{n-1}} + 1} \end{align*}which completes the proof.

Beautiful

Let $r_i=e^{a_i}$. Note that the condition says that all $a_i$ are nonnegative. Then define $f(x)=\frac{1}{e^x+1}$ and note that it's enough to show that $f(x)$ is convex on the interval $[0, \infty)$ and we will be done by Jensen's Inequality. Note that $$f''(x)=e^x(2(e^x+1)^{-3}e^x-(e^x+1)^{-2})\ge 0$$so we are done. $\blacksquare$

Define the $n$-variable polynomial $$P(r_1, r_2, \cdots, r_n) = \sum_{i=1}^n \frac 1{r_i+1} - \frac n{\sqrt[n]{r_1r_2\cdots r_n} + 1}.$$Suppose there existed some $r_i \neq r_j$, for instance $r_1$ and $r_2$ without loss of generality. I claim that $$P(r_1, r_2, \cdots, r_n) \geq P(\sqrt{r_1r_2}, \sqrt{r_1r_2}, \cdots, r_n).$$This is not difficult to prove because the second term and most of the first summation remains constant, and $$\frac 1{r_1+1} + \frac 1{r_2+1} = 1-\frac{r_1r_2-1}{r_1r_2+r_1+r_2+1} \geq 1-\frac{r_1r_2-1}{r_1r_2+2\sqrt{r_1r_2} + 1}$$by AM-GM and because $r_1r_2 \geq 1$, which is precisely what we wanted. Observe that $\sqrt{r_1r_2} \geq 1$ as well. The operation can be applied as long as the $r_i$ are not all equal. As a result, $$P(r_1, r_2, \cdots, r_n) \geq P(\sqrt[n]{r_1r_2\cdots r_n}, \sqrt[n]{r_1r_2\cdots r_n}, \cdots, \sqrt[n]{r_1r_2\cdots r_n}) = 0,$$so we are done.

We use Cauchy induction. We first prove the inequality for $n=2$. Let $r_1=a^2$ and $r_2=b^2$ so that $a,b \ge 1$. We have the inequality is $$\frac{1}{a^2+1}+\frac{1}{b^2+1} \ge \frac{2}{ab+1} \iff (ab+1)(a^2+b^2+2) \ge 2 (a^2+1)(b^2+1) \iff a^3b+ab^3-2a^2b^2 \ge a^2+b^2-2ab \iff (ab-1)(a-b)^2 \ge 0,$$which is true so we are done for $n=2$. We use standard induction to prove the inequality for $n=2^k$. The base case was proven above. Assume it is true for $n=2^k$. Then we have $$\frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n}+1} + \frac{1}{r_{n+1} + 1} + \frac{1}{r_{n+2} + 1} + \cdots +\frac{1}{r_{2n}+1} \ge \frac{n}{\sqrt[n]{r_1r_2 \cdots r_n} +1} + \frac{n}{\sqrt[n]{r_{n+1}r_{n+2} \cdots r_{2n}} +1} \ge \frac{2n}{\sqrt[2n]{r_1r_2 \cdots r_{2n}} +1},$$so we have finished the inequality. To finish the Cauchy induction, we will prove that the inequality being true for $n$ means that it is true for $n-1$. Taking $r_n=\sqrt[n-1]{r_1r_2 \cdots r_{n-1}}$, we have $$ \frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n-1}+1}+\frac{1}{\sqrt[n-1]{r_1r_2 \cdots r_{n-1}}+1} \ge \frac{n}{ \sqrt[n]{r_{1}r_{2} \cdots r_{n-1} \sqrt[n-1]{r_1r_2 \cdots r_{n-1}}}+1} \iff \frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n-1}+1} \ge \frac{n-1}{\sqrt[n-1]{r_1r_2 \cdots r_{n-1}} +1},$$which finishes the Cauchy induction so we are done.

Let $a_i=\ln r_i$ so that all $a_i$ are nonnegative, and let $$s=\frac{a_1+a_2\cdots + a_n}{n}.$$Then, we want to show $$\sum_{i=1}^n\frac{1}{e^{a_i}+1}\geq \frac{n}{e^s+1}.$$ Claim: $$f(x)=\frac{1}{e^x+1}$$is convex for $x\geq0$. This is because $$f''(x)=2(e^x+1)^{-3}e^{2x}-(e^x+1)^{-2}e^x=e^x\cdot \frac{e^x -1}{(e^x+1)^3},$$which is nonnegative when $x\geq0$. Therefore, by Jensen's, we are done.

OGGY_666 wrote: By Titu Inequality we have: LHS> n^2/(n+r1+...rn). .which is By applying AM-GM in the denominator> n^2/(n(r1...rn)^(1/n)+n) which is equivalent to: n/((r1...rn)^(1/n)+1)=RHS. Hence Proved. I almost had this solution initially, but it is a fakesolve since you flipped a sign (because you are AM-GMing the denominator so its actually <= in that step)

Define $a_i=\ln r_i,$ $a=\frac{1}{n}\sum_{k=1}^n a_k,$ $f(x)=\frac{1}{e^x+1}.$ By a straightforward check $f''(x)=\frac{e^x(e^{2x}-1)}{(e^x+1)^4},$ so by Jensen $$\sum_{k=1}^n \frac{1}{r_k+1} =\sum_{k=1}^n \frac{1}{e^{a_k}+1}\geq \frac{n}{e^a+1} =\frac{n}{\sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1} \text{ } \blacksquare$$

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Fix the product. We claim the left hand side is minimized when all the variables are equal. Assume not, then there are two variables $x\neq y$. It remains to prove \[ \frac{1}{x+1} + \frac{1}{y+1} \geq \frac{2}{\sqrt{xy}+1}. \]Expanding, it is equivalent to proving $(x+y+2)\sqrt{xy} \geq 2xy+x+y$, which can be factored as $(\sqrt{xy}-1)(\sqrt{x} - \sqrt{y})^2\geq 0$, which is obviously true. Then it remains to check the case when the variables are equal, which is trivial.

By Jensen's this is just asking us to prove that $f(x)=\frac{1}{e^x+1}$ is convex on $[0,\infty)$, which follows since its second derivative is $\frac{e^x(e^{2x}-1)}{(e^x+1)^4}$.

sqing wrote: orl wrote: Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. Prove that \[ \frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n}+1} \geq \frac{n}{ \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \] Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. For $\lambda\ge 1$, prove that \[ \frac{1}{\lambda r_{1} + 1} + \frac{1}{\lambda r_{2} + 1} + \cdots +\frac{1}{\lambda r_{n}+1} \geq \frac{n}{\lambda \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \] Substituting $r_1=\dfrac{p_1}{\lambda}$, we get that this isn't really a generalization because one can applicate the original inequality to $p_i$'s. We get $$\dfrac{2^k}{\sqrt[n]{p_1p_2\cdots p_n}+1}=\dfrac{2^k}{\lambda \sqrt{r_1r_2\cdots r_n}+1}$$While generalizing inequalities, these transformations should have been tried to check whether this is a real generalization or a fake one trivial by original problem.

Assume that some tuple $r_i \ge 1$ does not satisfy the inequality. We then show that for any real $x$, we can then create another tuple $t_i$ such that $\sum \frac{1}{r_i + 1} \ge \sum \frac{1}{t_i + 1}, \prod r_i = \prod t_i$, and $\mid \log \frac{\sqrt[n]{\prod r_i}}{r_j} \mid < x$ for all $j$. This is then a contradiction, since $\frac{1}{t_i + 1} \ge \sum \frac{n}{\sqrt[n]{\prod r_i}e^x + 1}$, for sufficiently small $x$ this can be as close to $\frac{n}{\sqrt[n]{\prod r_i} + 1}$ as we want, so for small enough $x$ we have $\frac{1}{t_i + 1} > \frac{1}{r_i + 1}$, contradiction. Let $r_1 \le r_2 \cdots \le r_n$, we show that replacing $r_1, r_n$ with $\sqrt{r_1r_n}$ reduces the value of the sum. Applying this $n$ times is guaranteed to reduce the maximal value of $\mid \log \frac{\sqrt[n]{\prod r_i}}{r_j} \mid$ by half , so applying this infinitely many times allows us to make said maximal value as small as we want. Of course, this is easy, we just desire $\frac{1}{a^2 + 1} + \frac{1}{b^2 + 1} \ge \frac{2}{ab + 1}$, expanding gives $(ab + 1)(a^2 + b^2 + 2) \ge 2a^2b^2 + 2a^2 + 2b^2 + 2$, which is equivalent to $a^3b + b^3a + 2ab \ge 2a^2b^2 + a^2 + b^2$, which is further equivalent to $ab(a - b)^2 \ge (a -b)^2$, which is obvious since $a,b \ge 1$.

Consider the function $f(x) = \frac{1}{e^x+1}$ for $x \geq 0.$ Clearly then $e^x \geq 1,$ and it is easy to see that $\frac{1}{1+x}$ is convex over $x \geq 1,$ so $f(x)$ is convex over $x \geq 0.$ Therefore, taking weights $a_1=a_2=\cdots=a_n = \frac{1}{n}$ and $x_i = \ln r_i,$ the inequality follows by Jensen's. QED