Prove that $\sum_{i=1}^n\frac{x_i}{\sqrt[nr]{x_i^{nr}+(n^{nr}-1)\prod_{j=1}^nx^r_j}} \ge 1$ for all $x_i>0$ and $r \geq \frac{1}{n}$.

For Bounty Hunters: This is "the current answer" I referred to in my bounty message. (I added this remark, so that people know which answer I meant if there are other answers.) The proof is cumbersome. I am very dissatisfied with it. I seek improvement.

Here is a proof of the OP's ineq for $r=\frac{k}{n}$ when $k$ is a positive integer. The proof the OP's conjecture for an arbitrary value $r\ge \frac1n$ is given later. I will show that for all positive integers $n$ and $k$, and for any real numbers $x_1,x_2,\ldots,x_n>0$, we have $$\sum_{i=1}^n\frac{x_i}{\sqrt[k]{x_i^k+(n^k-1)\prod_{j=1}^nx_j^{\frac{k}{n}}}}\ge1,\tag{0}$$ establishing the OP's conjecture for $r=\frac{k}{n}$. The proof is contingent upon the inequalities $$\frac{x_i}{\sqrt[k]{x_i^k+(n^k-1)\prod_{j=1}^nx_j^{\frac{k}{n}}}}\ge \frac{x_i^{1-\frac1{n^k}}}{\sum_{j=1}^n x_j^{1-\frac1{n^k}}}\tag{1}$$ for $i=1,2,\ldots,n$. The equality cases of $(0)$ and $(1)$ are the same:

  • $(n,k)=(2,1)$, and
  • $x_1=x_2=\ldots=x_n$.

To verify $(1)$, we assume wlog that $i=n$. We prove the equivalent inequality $$\left(1+\sum_{j=1}^{n-1}y_j^{1-\frac1{n^k}}\right)^k-1 \ge (n^k-1)\prod_{j=1}^{n-1}y_j^{\frac{k}{n}},\tag{2}$$ where $y_j=\frac{x_j}{x_n}$ for $j=1,2,\ldots,n-1$. There are five equality cases for $(2)$ if we allow $y_1,y_2,\ldots,y_{n-1}$ to be non-negative real numbers:

  • $n=1$,
  • $(n,k)=(2,1)$,
  • $y_1=y_2=\ldots=y_{n-1}=0$,
  • $y_1=y_2=\ldots=y_{n-1}=1$, and
  • $k=1$ with $y_1=y_2=\ldots=y_{n-1}$.

The lhs of $(2)$ is a sum of $(n^k-1)$ terms of the form $$Y_{(t_1,t_2,\ldots,t_{k})}=\prod_{\substack{1\le r\le k\\ t_r\ne n}} y_{t_r}^{1-\frac{1}{n^k}}$$ where $t_1,t_2,\ldots,t_{k}$ are positive integers not greater than $n$ s.t. not all of them are $n$. Write $t=(t_1,t_2,\ldots,t_k)$, and $T$ for the set of all possible tuples $t$.

Note that $|T|=n^k-1$. By AM-GM $$\left(1+\sum_{j=1}^{n-1}y_j^{1-\frac1{n^k}}\right)^k-1=\sum_{t\in T}Y_t\ge |T|\left(\prod_{t\in T}Y_t\right)^{1/|T|}=(n^k-1)\left(\prod_{t\in T} Y_t\right)^{\frac{1}{n^k-1}}.$$ Hence if we can show that $$\prod_{t\in T} Y_t=\left(\prod_{j=1}^{n-1} y_j^{\frac{k}{n}}\right)^{n^k-1},\tag{3}$$ then $(2)$ follows immediately. However by setting $y_n=1$, we have $$\prod_{t\in T}Y_t=\prod_{t_1=1}^n\prod_{t_2=1}^n\ldots\prod_{t_{k}=1}^n\prod_{r=1}^k y_{t_r}^{1-\frac1{n^k}}.\tag{4}$$

For $i=1,2,\ldots,n-1$, $y_i^{1-\frac{1}{n^k}}$ appears in the product in the rhs of $(4)$ in total \begin{align}\sum_{s=0}^k s\binom{k}{s}(n-1)^{k-s}&=k\sum_{s=1}^{k-1}\binom{k-1}{s-1}(n-1)^{(k-1)-(s-1)}\\&=k\big(1+(n-1)\big)^{k-1}=kn^{k-1}\end{align} times. This means $$\prod_{t\in T}Y_t=\prod_{i=1}^n \left(y_i^{1-\frac{1}{n^k}}\right)^{kn^{k-1}}=\prod_{i=1}^n\left( y_i^{\frac{k}{n}}\right)^{n^k-1}.$$ This justifies $(3)$ and we are done.

Here is a proof of the OP's ineq for any real value $r\ge \frac1n$. We want to show that $(2)$ is true for any real number $k\ge 1$ and for any real numbers $y_1,y_2,\ldots,y_{n-1}\ge 0$. Furthermore when $0<k<1$ and $n>1$, it is not difficult to see that that there are always counterexamples of $(2)$. Hence, when $k>0$ and $n>1$, the inequalities $(0)$ and $(1)$ always hold for any positive reals $x_1,x_2,\ldots,x_n$ if and only if $k\ge 1$; the ineq $(2)$ always hold for any non-negative reals $y_1,y_2,\ldots,y_{n-1}$ iff $k\ge 1$. This means: for $r>0$ and $n>1$, the OP's inequality holds for any positive reals $x_1,x_2,\ldots,x_n$ if and only if $r\ge \frac1n$.

Rewrite $(2)$ in the following form: $$\left(1+\sum_{j=1}^m z_j\right)^k -1 \ge \big((1+m)^k-1\big)\prod_{j=1}^m z_j^{\frac{k (1+m)^{k-1}}{(1+m)^k-1}},$$ where $z_j=y_j^{1-\frac{1}{n^k}}$ and $m=n-1$. By setting $z_j=\frac{w_j}{m}$ where $w_1,w_2,\ldots,w_m\ge 0$, we get yet another form of $(2)$. Let $A_m$ and $G_m$ denote the arithmetic mean and the geometric mean of $w_1,w_2,\ldots,w_m$. Then we need to prove $$(1+A_m)^k-1\geq \frac{(1+m)^k-1}{m^{\frac{km(1+m)^{k-1}}{(1+m)^k-1}}}G_m^{\frac{km(1+m)^{k-1}}{(1+m)^k-1}}$$ for all real numbers $k\ge 1$. Since $A_m\ge G_m$, it suffices to show that$$(1+G_m)^k-1\geq \frac{(1+m)^k-1}{m^{\frac{km(1+m)^{k-1}}{(1+m)^k-1}}}G_m^{\frac{km(1+m)^{k-1}}{(1+m)^k-1}}$$ for all real numbers $k\ge 1$. More generally, we want to prove $$\frac{(1+g)^k-1}{(1+\mu)^k-1} \geq \left(\frac{g}{\mu}\right)^{\frac{k\mu(1+\mu)^{k-1}}{(1+\mu)^k-1}}\tag{5}$$ for all real numbers $g$, $\mu$, and $k$ such that $g\ge 0$, $\mu> 0$, and $k\ge 1$ (with equality cases $g=0$, $g=\mu$, and $k=1$). We will also see that $(5)$ is reversed when $0<k<1$ and $k<0$ (in both ranges, the equality conditions are $g=0$ and $g=\mu$).

Let $$F(g)=\left\{\begin{array}{ll} (1+g)^k(1+\mu)^k-\mu g\left(\frac{(1+g)^k-(1+\mu)^k}{g-\mu}\right)-\frac{g(1+g)^k-\mu(1+\mu)^k}{g-\mu} &\text{if }g\ne \mu,\\ (1+\mu)^{2k}-(1+\mu)^{k}(1+k\mu) & \text{if }g=\mu.\end{array}\right.$$ Then $$H'(g)=\frac{k (g-\mu) F(g)}{g^{\frac{k\mu(1+\mu)^{k-1}}{(1+\mu)^k-1}+1}\big((1+\mu)^k-1\big)(1+\mu)(1+g)},$$ if $$H(g)=\frac{(1+g)^k-1}{g^{\frac{k\mu(1+\mu)^{k-1}}{(1+\mu)^k-1}}}.$$ If we can show that when $k>1$, $F(g)>0$ for all $g>0$, then it follows that $H(g)$ achieves the minimum value at $g=\mu$, proving $(5)$. Anyway, $F(g)>0$ for all $g>0$ if and only if $$(1+g)^{k-1}(1+\mu)^{k-1}\ge \frac{g(1+g)^{k-1}-\mu(1+\mu)^{k-1}}{g-\mu}\tag{6}$$ for all $g\ne \mu$ and $k\ge 1$ (with equality case $k=1$). To prove $(6)$, it suffices to assume that $g>\mu\ge 0$.

In fact, we will prove that $(6)$ is true for all positive reals $g,\mu$ s.t. $g\ne \mu$ and for any $k\in(-\infty,0]\cup [1,\infty)$ with equality cases $k=0$ and $k=1$. Moreover $(6)$ is reversed when $k\in(0,1)$ without equality cases. The reversed inequality of $(6)$ when $0<k<1$ is significant because it allows us to find counterexamples of $(0)$, $(1)$, and $(2)$ when $n>1$ and $0<k<1$.

If $\kappa=k-1$, then $(6)$ is equivalent to $$(1+g)^{\kappa}\frac{(1+\mu)^{\kappa}-1}{\mu}\ge \frac{(1+g)^{\kappa}-(1+\mu)^{\kappa}}{g-\mu}\tag{7}$$ for all $g>\mu>0$ and $\kappa \ge 0$ and $\kappa\le -1$ (with equality cases $\kappa=0$ and $\kappa=-1$). We have a reversed version of $(7)$ for $-1<\kappa<0$ (without equality cases). Once $(7)$ is established, $(5)$ and $(6)$ follow immediately. Hence $(0)$, $(1)$, and $(2)$ are true for all real numbers $k\ge 1$.

Now we prove $(7)$ for $\kappa\ge 0$ and $\kappa<-1$, as well as its reversed version for $-1\le \kappa<0$. The main idea is Bernoulli's ineq: $$(1+x)^{\alpha}\ge 1+\alpha x$$ for all $x>-1$ and $\alpha\in (-\infty,0]\cup[1,\infty)$ (with equality cases $x=0$, $\alpha=0$, and $\alpha=1$). The inequality above is reversed for $0<\alpha<1$ (with $x=0$ as the sole equality condition). We assume $g>\mu>0$ throughout.

If $\kappa \ge 1$, then by Bernoulli's ineq $$\frac{(1+\mu)^\kappa-1}\mu\ge \kappa$$ and $$\frac{1-\left(\frac{1+\mu}{1+g}\right)^\kappa}{g-\mu}=\frac{1-\left(1-\frac{g-\mu}{1+g}\right)^\kappa}{g-\mu}\le \frac{\kappa}{1+g}.$$ Thus $$\frac{(1+\mu)^\kappa-1}\mu\ge \kappa\ge\frac{\kappa}{1+g}\ge\frac{1-\left(\frac{1+\mu}{1+g}\right)^\kappa}{g-\mu}$$ proving $(7)$. For $\kappa\ge 1$, $(7)$ is a strict ineq.

If $0\le \kappa<1$, then Bernoulli's ineq implies $$\frac{1-\left(\frac{1}{1+\mu}\right)^\kappa}{\mu}=\frac{1-\left(1-\frac{\mu}{1+\mu}\right)^\kappa}{\mu}\ge \frac{\kappa}{1+\mu}$$ and $$\frac{\left(\frac{1+g}{1+\mu}\right)^\kappa-1}{g-\mu}=\frac{\left(1+\frac{g-\mu}{1+\mu}\right)^\kappa-1}{g-\mu}\le \frac{\kappa}{1+\mu}.$$ Hence $$(1+g)^\kappa\frac{1-\left(\frac{1}{1+\mu}\right)^\kappa}{\mu}\ge \frac{1-\left(\frac{1}{1+\mu}\right)^\kappa}{\mu} \ge \frac{\kappa}{1+\mu}\ge \frac{\left(\frac{1+g}{1+\mu}\right)^\kappa-1}{g-\mu}$$ proving $(7)$. For $0<\kappa<1$, $(7)$ is a strict ineq.

We also show that the inequality $(7)$ is flipped when $-1\le\kappa<0$. For $-1\le \kappa<0$, Bernoulli's ineq implies $$\frac{\left(\frac{1}{1+\mu}\right)^{-\kappa}-1}{\mu}=\frac{\left(1-\frac{\mu}{1+\mu}\right)^{-\kappa}-1}{\mu}\le\frac{\kappa}{1+\mu}$$ and $$\frac{1-\left(\frac{1+g}{1+\mu}\right)^{-\kappa}}{g-\mu}=\frac{1-\left(1+\frac{g-\mu}{1+\mu}\right)^{-\kappa}}{g-\mu}\ge\frac{\kappa}{1+\mu}.$$ Thus $$\frac{\left(\frac{1}{1+\mu}\right)^{-\kappa}-1}{\mu}\le\frac{\kappa}{1+\mu}\le\frac{1-\left(\frac{1+g}{1+\mu}\right)^{-\kappa}}{g-\mu}$$ reversing $(7)$. The reversed version of $(7)$ is strict for $-1<\kappa<0$.

Lastly we want to prove that for $\kappa<-1$, $(7)$ holds once again. Using Bernoulli's ineq, we get $$\frac{\left(\frac{1}{1+\mu}\right)^{-\kappa}-1}{\mu}=\frac{\left(1-\frac{\mu}{1+\mu}\right)^{-\kappa}-1}{\mu}\ge\frac{\kappa}{1+\mu}$$ and $$\frac{1-\left(\frac{1+g}{1+\mu}\right)^{-\kappa}}{g-\mu}=\frac{1-\left(1+\frac{g-\mu}{1+\mu}\right)^{-\kappa}}{g-\mu}\le\frac{\kappa}{1+\mu}.$$ Thus $$\frac{\left(\frac{1}{1+\mu}\right)^{-\kappa}-1}{\mu}\ge\frac{\kappa}{1+\mu}\ge\frac{1-\left(\frac{1+g}{1+\mu}\right)^{-\kappa}}{g-\mu}$$ establishing $(7)$. The ineq $(7)$ is strict for $\kappa<-1$.

From the result above, we can also show that $$\left(1+\sum_{j=1}^{n-1}y_j^{\frac{n}{n-1}\left(1-\frac1{n^k}\right)}\right)^k-1\ge (n^k-1)\prod_{j=1}^{n-1}y_j^{\frac{k}{n-1}}$$ for every integer $n>1$, for any real number $k\ge 1$, and for all real numbers $y_1,y_2,\ldots,y_{n-1}\ge 0$ with two equality cases: $(n,k)=(2,1)$ and $y_1=y_2=\ldots=y_{n-1}\in\{0,1\}$. This implies $$\frac{x_i}{\sqrt[k]{x_i^k+(n^k-1)\prod_{j\ne i}x_j^{\frac{k}{n-1}}}}\ge \frac{x_i^{\frac{n}{n-1}\left(1-\frac1{n^k}\right)}}{\sum_{j=1}^nx_j^{\frac{n}{n-1}\left(1-\frac1{n^k}\right)}}$$ and $$\sum_{i=1}^n\frac{x_i}{\sqrt[k]{x_i^k+(n^k-1)\prod_{j\ne i}x_j^{\frac{k}{n-1}}}}\ge 1$$ for every integer $n>1$, for any real number $k\ge 1$, and for all real numbers $x_1,x_2,\ldots,x_{n}> 0$. The equality conditions of the last two inequalities are $(n,k)=(2,1)$ or $x_1=x_2=\ldots=x_n$. The second problem of IMO 2001 is a special case of the last inequality, where $(n,k)=(3,2)$.


The following reasoning can give a solution.

Let $nr=\alpha$.

Thus, $\alpha\geq1$ and we need to prove that: $$\sum_{i=1}^n\frac{x_i}{\left(x_i^{\alpha}+(n^{\alpha}-1)\prod\limits_{j=1}^nx_j^{\frac{\alpha}{n}}\right)^{\frac{1}{\alpha}}}\geq1.$$ Now, by Holder $$\left(\sum_{i=1}^n\frac{x_i}{\left(x_i^{\alpha}+(n^{\alpha}-1)\prod\limits_{j=1}^nx_j^{\frac{\alpha}{n}}\right)^{\frac{1}{\alpha}}}\right)^{\alpha}\sum_{i=1}^nx_i\left(x_i^{\alpha}+(n^{\alpha}-1)\prod\limits_{j=1}^nx_j^{\frac{\alpha}{n}}\right)\geq\left(\sum_{i=1}^nx_i\right)^{\alpha+1}.$$ Thus, it's enough to prove that: $$\left(\sum_{i=1}^nx_i\right)^{\alpha+1}\geq\sum_{i=1}^nx_i^{\alpha+1}+(n^{\alpha}-1)\prod\limits_{j=1}^nx_j^{\frac{\alpha}{n}}\sum_{i=1}^nx_i,$$ which is true by Muirhead for any integer $\alpha\geq n$.

Now, let $\sum\limits_{i=1}^nx_i$ be a constant and $\sum\limits_{i=1}^nx_i^{\alpha+1}$ be a constant.

Thus, by the Vasc's EV Method (https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf Corollary 1.8 b )

$\prod\limits_{j=1}^nx_j$ gets a maximal value for equality case of $n-1$ variables.

Since the last inequality is homogeneous, it's enough to assume $x_2=...=x_n=1$ and $x_1=x$,

which gives something easier.

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Inequality

Summation

Conjectures

Algebra Precalculus

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