# If $\sum\limits_{i=1}^{n+1} \frac{1}{ny_i+1}=1$, then $\prod\limits_{i=1}^{n+1}y_i\geq 1$?

Let $y_1,\ldots,y_{n+1}$ be positive real numbers satisfying $\displaystyle{\sum_{i=1}^{n+1} \frac{1}{ny_i+1}=1}$.

Is it true that $y_1y_2\cdots y_{n+1}\geq 1$?

Added: can we determine this inequality in terms of high-school math? (e.g. Cauchy-Schwarz inequality, arithmetic mean-geometric mean inequality)

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Could you please consider using somewhat more descriptive titles? I'm just guessing you're the same person as the "Mark Jin" who asked this question. Also, what do you want to achieve with these inequalities? No offense, I'm just curious. –  t.b. Jun 23 '11 at 4:28
I'm same person. I just want to prove $\sum_{i=1}^{n+1}\frac{1}{ny_i+1}=1$ implies $y_1\cdots y_{n+1}\geq 1$. From some Cauchy-Schwarz inequality, I can get $y_1+\cdots y_{n+1}\geq n+1$ and $1/y_1+1/y_2\cdots/y_{n+1}\geq n+1$. That's why I asked that question yesterday. And I'm unregistered user hence I should use another name. –  Mark Jin Jun 23 '11 at 4:32
@Jonas: I would prefer that \displaystyle was not used in question titles. It is still typeset as large inline text, and on the front page I feel it looks rather conspicuous. –  Rahul Jun 23 '11 at 4:58
@Rahul: I agree with that. Is this better, or still conspicuous? –  t.b. Jun 23 '11 at 5:05
@Mark: please register your account. You won't have any trouble with logging back in that way. –  Qiaochu Yuan Jun 23 '11 at 12:15

You could have a look at Lagrange multiplier method.

Edit Here is a proof using only convexity and basic calculus.

First case: $y_i\ge1/n$ for every $i$

Consider the function $u$ defined by $$u(t)=\frac1{1+n\mathrm{e}^t}.$$ Computing the second derivative of $u$, one sees that $u$ is convex on the domain where $n\mathrm{e}^t\ge1$. Hence, if $n+1$ numbers $t_i$ are such that $n\mathrm{e}^{t_i}\ge1$ for every $i$, then $$u(t_1)+\cdots+u(t_{n+1})\ge(n+1)u\left(\frac{t_1+\cdots+t_{n+1}}{n+1}\right).$$ Assume that $y_i\ge1/n$ for every $i$ and apply the inequality above to $t_i=\log y_i$, then the LHS is by hypothesis $1$ and $t_1+\cdots+t_{n+1}=\log(y_1\cdots y_{n+1})$ hence the RHS is $$(n+1)u\left(\frac{\log(y_1\cdots y_{n+1})}{n+1}\right).$$ Since $(n+1)u(t)\le1$ if and only if $t\ge0$, this shows that $\log(y_1\cdots y_{n+1})\ge0$ and we are done.

Second case: $y_i<1/n$ for some $i$

Note that this can happen at most for one index $i$ and assume for example that $y_1<1/n$, then $y_i\ge1/n$ for every $i\ne1$ and $y_i>1/n$ for at least one index $i\ne1$. Assume for instance that $y_2>1/n$. Define a deformation $(y_1(z),y_2(z))$ of $(y_1,y_2)$ for every small enough nonnegative $z$ by $y_1(z)=y_1+z$ and $$\frac1{1+ny_1(z)}+\frac1{1+ny_2(z)}=\frac1{1+ny_1}+\frac1{1+ny_2}.$$ Then the product $y_1(z)y_2(z)$ is a decreasing function of $z$ as long as $y_1(z)\le1/n\le y_2(z)$ (the proof is in the addendum below) hence the result holds for $z=0$ as soon as it holds for such a given positive $z$.

If this is enough to move the value of $y_1$ up to $1/n$, the proof is complete. Otherwise this means that $y_1(z)<1/n=y_2(z)$ for a given $z$. Apply the same procedure to this new $y_1$ and to another $y_i$ such that $y_i>1/n$. After at most $n$ steps, one gets a collection $(\bar y_i)$ which still satisfies the hypothesis of the post and such that $\bar y_i\ge1/n$ for every $i$. Furthermore, $$y_1\cdots y_{n+1}\ge \bar y_1\cdots \bar y_{n+1}.$$ The first case shows that $\bar y_1\cdots \bar y_{n+1}\ge1$, hence we are done.

Addendum: the function $z\mapsto y_1(z)y_2(z)$ is decreasing

Differentiating, one sees that one should show that $$y_2(z)(1+ny_1(z))^2< y_1(z)(1+ny_2(z))^2.$$ Using the fact that $y_2(z)> y_1(z)$, this is equivalent to $n^2y_1(z)y_2(z)> 1$, which in turn is equivalent to $$\frac1{1+ny_1(z)}+\frac1{1+ny_2(z)}< 1,$$ hence the assertion holds.

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I want use only high-school level method. (e.g. Cauchy-Schwarz inequality) –  Mark Jin Jun 23 '11 at 6:21

Let $a_i = \frac{1}{ny_i + 1}$, then $\sum a_i = 1$, $0 < a_i < 1$ for all $i$, and you want to show that

$$\prod_i \left(\frac{1}{a_i} - 1\right) \ge n^{n+1}$$

Clear denominator, and use the sum condition, you want to show

$$\prod_i (a_1 + \cdots + \hat{a_i} + \cdots + a_{n+1}) \ge n^{n+1} \prod a_i$$

(the hat means I ignore that term in summation) This follows from applying AM-GM for each thing on the left.

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Nice.    –  Did Jun 23 '11 at 16:32

In the case $n=1$, check it yourself that you get $y_1y_2=1$.

In the case $n=2$ the relation translates to $8y_1y_2y_3=2(y_1+y_2+y_3)+2 \geq 6 \sqrt[3]{y_1y_2y_3}+2$. Denote $\sqrt[3]{y_1y_2y_3}=k$.

We have $8k^3-6k-2 \geq 0$ which is equivalent to $(k-1)(8k^2+8k+2) \geq 0$. Since the second factor is positive for $k>0$ it follows that $k \geq 1$.

Maybe something similar can be tried for $n$ greater than $4$.

 The same reasoning, with a greater deal of computations works for $n=4$ too.

Lagrange multipliers method gives us that the optimal value for the product is obtained for equal values of $y_i$, and those values are equal to $1$. Therefore the inequality $y_1...y_{n+1}\geq 1$ is true.

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