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 $x_i$.)

  • 2
    $\begingroup$ The claim is untrue. Take $n=2,x_1=1,x_2=2$ and $r\in(0,0.5)$. $\endgroup$ – Jam Jan 29 '20 at 18:41
  • $\begingroup$ @Jam Hmm... what if $r\geq 1$ ? $\endgroup$ – Hulkster Jan 29 '20 at 18:44
  • $\begingroup$ Seems to work if $r\ge 1$. I've not proven it yet though. $\endgroup$ – 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 $0<k<1$ 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 $(n,k)=(3,2)$.

  • $\begingroup$ Thanks for you efforts! $\endgroup$ – 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.


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.