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.