# 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 $r \geq \frac{1}{n}$.

Prove that, for all $$x_1,x_2,\ldots,x_n>0$$ and $$r \geq \frac{1}{n}$$, it holds that $$\sum_{i=1}^n\frac{x_i}{\sqrt[nr]{x_i^{nr}+(n^{nr}-1)\prod \limits_{j=1}^nx^r_j}} \ge 1.$$

This is a slightly modified version of my earlier question here (which corresponds to the case $$r=1$$).

The case $$n=3$$ and $$r=\frac{2}{3}$$ can be reduced to the inequality problem of IMO 2001 (Problem 2).

The case $$r=\frac{1}{n}$$ can be reduced to $$\sum_{i=1}^n\frac{x_i^{n-1}}{x_i^{n-1}+(n-1) \prod_{j\neq i}x_j}\geq \sum_{i=1}^n\,\frac{x_i^{n-1}}{x_i^{n-1}+\sum_{j\neq i}\,x_j^{n-1}}=1\,.$$
(Note that we have an equality case when $$n=2$$ and $$r=\frac1n$$ here, regardless of the values of the variables $$x_i$$.)

• The claim is untrue. Take $n=2,x_1=1,x_2=2$ and $r\in(0,0.5)$. – Jam Jan 29 '20 at 18:41
• @Jam Hmm... what if $r\geq 1$ ? – Hulkster Jan 29 '20 at 18:44
• Seems to work if $r\ge 1$. I've not proven it yet though. – Jam Jan 29 '20 at 18:50

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 $$0 and $$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 $$0 and $$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 $$0 is significant because it allows us to find counterexamples of $$(0)$$, $$(1)$$, and $$(2)$$ when $$n>1$$ and $$0.

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 $$(n,k)=(3,2)$$.

• Thanks for you efforts! – Hulkster Apr 22 '20 at 23:39

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.