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)
 A: 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.
A: 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.
A: 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 $4$. 
[edit] 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.
