Let $b_i=\tan(a_i)$. We are given $\sum_{i=0}^n \frac{b_i-1}{b_i+1} \geq {n-1}$. Rearranging this gives $\sum_{i=0}^n \frac{1}{b_i+1} \leq 1$ Let $c_i=\frac{1}{b_i+1}$, so $0<c_i<1$. We have $\sum \frac{1}{b_i} = \sum \frac{c_i}{1-c_i}=\sum (1-\frac{1}{1-c_i})$. This last function is concave on $(0,1)$, so it follows that for a fixed $\sum c_i$ that $\sum \frac{1}{b_i}$ is maximized when all the $c_i$ are equal. Since increasing a $c_i$ decreases the corresponding $b_i$ and thus increases $1/{b_i}$, it follows that $\sum \frac{1}{b_i}$ is maximized when all the $c_i$ are equal to $\frac{1}{n+1}$, i.e. \[ \sum_{i=0}^n \frac{1}{b_i} \leq \frac{n+1}{n} \] This means the harmonic mean of the $b_i$ is at least $n$, so the geometric mean must also be at least $n$.

kevinatcausa wrote: $\sum \frac{1}{b_i} = \sum \frac{c_i}{1-c_i}=\sum (1-\frac{1}{1-c_i})$. This last function is concave on $(0,1)$... $\frac{c_i}{1-c_i}=\frac{1}{1-c_i} -1$, not sure whether your argument still works.

Solution with smoothing! Rearrange in the obvious way, then let $ b_{i} = \tan{a_{i}}$. Lemma: When $ b_{i},b - {j}\ge1$, we have $ \dfrac{2}{1 + \sqrt {b_{i}b_{j}}}\le\dfrac{1}{1 + b_{i}} + \dfrac{1}{1 + b_{j}}$. Proof: Expand, it becomes $ (b_{i} - b_{j})^{2}(b_{i}b_{j} - 1)\ge0$. What this means is, if we have $ b_{i}b_{j}\ge1$, we can change them into $ \sqrt {b_{i}b_{j}},\sqrt {b_{i}b_{j}}$ so that the product of the $ b_{i}$'s remains the same and the $ b_{i}$ still satisfy the $ \displaystyle\sum_{i=0}^{n}\dfrac{1}{1+b_{i}}\le1$ condition, and furthermore as long as we can find two terms whose product is at least 1, we can make everything equal, hence reducing the problem to the equality case. Now we can prove that we can always find two terms whose product is at least 1. If all of the $ b_{i}$ are at least 1 to start with, we're done; note that if $ b_{i}b_{j}\ge1$, the square root is also at least 1. Now, if more than two of the $ b_{i}$ are less than 1, we find that the sum must be more than 1 (because their corresponding terms will be more than 1/2), contradiction. So only one of them can be 1. Say $ b_{0} < 1$. We claim that there's another term $ b_{i}$ with $ b_{0}b_{i}\ge1$. Assume the opposite, so $ b_{i} < 1/b_{0}$ for all $ 1\le i\le n$. Then, $ \displaystyle\sum_{i = 0}^{n}\dfrac{1}{1 + b_{i}} = \dfrac{1}{1 + b_{0}} + \displaystyle\sum_{i = 0}^{n}\dfrac{1}{1 + b_{i}} > \dfrac{1}{1 + b_{0}} + n\cdot\dfrac{b_{0}}{1 + b_{0}}\ge1$, contradiction. Thus, we can operate on $ b_{0}$ and $ b_{i}$, making every term greater than 1, and proceed as before. Now it's left to consider $ b_{0} = b_{1} = \cdots = b_{n}$, but this is obvious.

It was brought to my attention that there's a hole in my above proof - we have to prove that we can indeed keep operating until all the $ b_{i}$ are equal. This actually isn't true, but we can prove, intuitively, that they get closer and closer together. Here's an outline (I haven't actually worked out all the details). Define Let $ g$ be the Geometric Mean of all the terms, and define $ c_{i}$ to be $ g/b_{i}$ when this is at least 1, or the reciprocal when it's not, so $ c_{i} > 1$. Note that if the terms are not all equal, we can find two terms, on on each side of $ g$. It's not too hard to show that after operating on these, the product of the $ c_{i}$ goes down. The limit as the number of operations goes to infinity should be 1 (which I don't have a rigorous proof for yet, I'll post one if/when I find it). This means that $ \text{max}b_{i}/\text{min}b_{i}$ must to 1 as well. Using the $ \displaystyle\sum\dfrac{1}{1 + b_{i}}$ condition, we can get that $ \text{max}b_{i}\ge n$. Assume the product of the $ b_{i}$'s is less than $ n^{n + 1}$, letting it be $ n^{n + 1} - x$, say. But note that the product is at least $ (\text{min}b_{i})^{n + 1}$, which, since max/min goes to 1, will surpass $ n^{n + 1} - x$ after a sufficiently large number of operations. Again, I still have to show that the limit is 1...hopefully I'll get back to this.

OK I think this works. The limit clearly exists because the product is strictly decreasing and always stays above 1. Let the limit be L. Now, we can show that the product always decreases by a factor of $ c_{i}$ or $ c_{j}$ depending on which side of $ g$ $ b_{i}b_{j}$ is on. Thus, we have $ (L+\epsilon)c_{i}>L$, and we find that $ L/(L+\epsilon)<c_{i}$. Thus, as $ \epsilon$ goes to 0, $ c_{i}$ goes to 1, and thus the product of the $ c_{i}$'s goes to 1.

kevinatcausa wrote: Let $b_i=\tan(a_i)$. We are given $\sum_{i=0}^n \frac{b_i-1}{b_i+1} \geq {n-1}$. Rearranging this gives $\sum_{i=0}^n \frac{1}{b_i+1} \leq 1$ Let $c_i=\frac{1}{b_i+1}$, so $0<c_i<1$. Adopting this notation, another way to finish this by smoothing is to see that if we have $\sum{c_i}=k\le1$, then we wish to minimize $P=\prod{b_i}=\prod\left(\frac{1}{c_i}-1\right)$. But we can show that if some $u,v$ exist with $c_u<k/(n+1)<c_v$, then changing $(c_u,c_v)$ to $(k/(n+1),c_u+c_v-k/(n+1))$ will not change the sum of the $c_i$'s but will decrease the desired product, since \begin{align*} \left(\frac{1}{c_u}-1\right)\left(\frac{1}{c_v}-1\right) &> \left(\frac{1}{\frac{k}{n+1}}-1\right)\left(\frac{1}{c_u+c_v-\frac{k}{n+1}}-1\right)\\ &= \left(\frac{n+1}{k}-1\right)\left(\frac{n+1}{(n+1)(c_u+c_v)-k}-1\right)\\ \Longleftrightarrow\frac{1-c_u-c_v}{c_uc_v} &> \frac{(n+1)^2}{k[(n+1)(c_u+c_v)-k]}-\frac{n+1}{k}-\frac{n+1}{(n+1)(c_u+c_v)-k)}\\ &=\frac{(n+1)^2}{k[(n+1)(c_u+c_v)-k]}-\frac{(n+1)^2(c_u+c_v)}{k[(n+1)(c_u+c_v)-k]}\\ &=\frac{(n+1)^2(1-c_u-c_v)}{k[(n+1)(c_u+c_v)-k]}. \end{align*}But $c_u+c_v<\sum{c_i}\le1$, so this is equivalent to \[\frac{1}{c_uc_v} > \frac{(n+1)^2}{k[(n+1)(c_u+c_v)-k]},\]and because $(n+1)(c_u+c_v)>(n+1)c_v>k$, we can expand and rearrange to get\[0>k^2-k(n+1)(c_u+c_v)+c_uc_v(n+1)^2=(k-c_u(n+1))(k-c_v(n+1)),\]which is true because $c_u<k/(n+1)<c_v$ by assumption. Since this smoothing process brings more of the $c_i$ to $k/(n+1)$ each time, it clearly terminates and minimizes $P$ when all $c_i=k/(n+1)$, so noting that $k\le1$, \[P=\prod\left(\frac{1}{c_i}-1\right)\ge\left(\frac{n+1-k}{k}\right)^{n+1}\ge n^{n+1},\]as desired. Edit: Darn... so evidently \[P=\prod{b_i}=\prod\left(\frac{1}{c_i}-1\right)\ge\prod\left(\frac{c_0+\cdots+c_{i-1}+c_{i+1}+\cdots+c_n}{c_i}\right)\ge n^{n+1}\]by AM-GM.

Another way to smooth is to note that if not all $b_i$ are equal, then we can find $b_m$ greater and $b_n$ less than the geometric mean of all $b_i$. Then, we prove that if $a<b$ and $e>0$, then $\frac{1}{a+1} + \frac{1}{b+1} > \frac{1}{ae+1} + \frac{1}{\frac{a}{e}+1}$ by expanding, collecting terms, blahblahblah. Then, at each step, we can choose $e$ so that either $b_me$ or $\frac{b_n}{e}$ is equal to the geometric mean of all $b_i$, so we eventually get to the equality case. Although the "infinite" smoothing by replacing two terms with their geometric mean is intuitive, it requires some limit stuff to rigorize, which may be frowned upon...

MithsApprentice wrote: Let $a_0,a_1,\cdots ,a_n$ be numbers from the interval $(0,\pi/2)$ such that \[ \tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})+\cdots +\tan (a_n-\frac{\pi}{4})\geq n-1. \] Prove that \[ \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}. \] What about AM-GM? Let $x_i=\tan a_i$ and rewrite the given condition into $\frac 1{x_0+1}+\frac1{x_1+1}+\cdots+\frac 1{x_n+1}\leq 1.$ We have to show that $x_0x_1x_2\cdots x_n\geq n^{n+1}.$ Note that using the given condition, we have \[\frac{x_0}{x_0+1}\geq \frac{1}{x_1+1}+\frac 1{x_2+1}+\cdots+\frac 1{x_n+1}\geq n\sqrt[n]{\frac 1{(x_1+1)(x_2+1)\cdots (x_n+1)}}.\] Getting its analogues and multiplying all the relations together, we get \[x_0x_1x_2\cdots x_n\geq n^{n+1}.\] Equality occurs iff $x_i=n\forall i=0,1,\cdots n.\Box$

Sorry for reviving an old topic, using Jensen's inequality on the function $ln(\frac{1+x}{1-x})$, but it is concave on $(-1,0]$ and convex on $[0,1)$. How does smoothing solve this problem so we can use Jensen's inequality here? Thank you!

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=266622

Solution from Twitch Solves ISL: Let $x_i = \tan(a_i - \frac{\pi}{4})$. Then we have that \[ \tan a_i = \tan(a_i - 45^{\circ} + 45^{\circ}) = \frac{x_i + 1}{1 - x_i}. \]If we further substitute $y_i = \frac{1 - x_i}{2} \in (0,1)$, then we have to prove that the following statement: Claim: If $\sum_0^n y_i \le 1$ and $y_i \ge 0$, we have \[ \prod_{i=1}^n \left( \frac{1}{y_i}-1 \right) \ge n^{n+1}. \]Proof. Homogenizing, we have to prove that \[ \prod_{i=1}^n \left( \frac{y_0+y_1+y_2+\dots+y_n}{y_i}-1 \right) \ge n^{n+1}. \]By AM-GM, we have \[ \frac{y_1+y_2+y_3+\dots+y_n}{y_0} \ge n \sqrt[n]{\frac{y_1y_2y_3\dots y_n}{y_1}}. \]Cyclic product works. $\blacksquare$ Remark: Alternatively, the function $x \mapsto \log(1/x-1)$ is a convex function on $(0,1)$ so Jensen inequality should also work.

\begin{align*} & \tan(a_i - \dfrac{\pi}{4})\\ &= \dfrac{\tan a_i - \tan \dfrac{\pi}{4}}{1+ \tan a_i \tan \dfrac{\pi}{4}}\\ &= \dfrac{\tan a_i - 1}{\tan a_i + 1} \end{align*} Let $x_i = \tan a_i \forall i$, So $$\tan(a_i - \dfrac{\pi}{4}) = \dfrac{x_i - 1}{x_i + 1}$$ Since $a_i \in (0,\dfrac{\pi}{4}) \forall i$, we must have $x_i > 0 \forall i$. Then we have $-1 < \tan(a_i - \dfrac{\pi}{4}) < 1 \forall i$. We now apply a local adjustment. We want to minimise $xy$ given a fixed $\dfrac{x-1}{x+1} + \dfrac{y-1}{y+1} = c$. Let us denote $x' = \dfrac{x-1}{x+1}$, Thus $x= \dfrac{1+x'}{1-x'}$, $x'+y'=c$. \begin{align*} &xy\\ &= \dfrac{(1+x')(1+c-x')}{(1-x')(1-c+x')}\\ &= \dfrac{-x'^2 + cx' + c + 1}{-x'^2 + cx' - c + 1} \end{align*}Which reaches minimum at $x'= \dfrac{c}{2}$, Where $x'=y'$. Consider the following operation. Each time we take the 2 $x_i'$s with the greatest difference, And set them to be equal to their arithmetic mean. This decreases the product of all $x_i$s as shown. Each time this is done, The maximum difference between 2 $x_i'$s decrease So eventually we have $$x_i' = x_j' = \dfrac{n-1}{n+1} \forall i,j \in \mathbb{Z_0^+}, i,j \geq 0$$ We then evaluate the required product of all $x_i$s under this worst case. For any $0 \leq i \leq n$, \begin{align*} &x_i\\ &= \dfrac{1+ x_i'}{1- x_i'}\\ &= \dfrac{1+ \dfrac{n-1}{n+1}}{1- \dfrac{n-1}{n+1}}\\ &= \dfrac{n+1+n-1}{n+1-n+1}\\ &=n \end{align*} Thus we conclude $$ x_0x_1 \dots x_n \geq n^{n+1} \forall n$$$\square$

Potla wrote: MithsApprentice wrote: Let $a_0,a_1,\cdots ,a_n$ be numbers from the interval $(0,\pi/2)$ such that \[ \tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})+\cdots +\tan (a_n-\frac{\pi}{4})\geq n-1. \]Prove that \[ \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}. \] What about AM-GM? Let $x_i=\tan a_i$ and rewrite the given condition into $\frac 1{x_0+1}+\frac1{x_1+1}+\cdots+\frac 1{x_n+1}\leq 1.$ We have to show that $x_0x_1x_2\cdots x_n\geq n^{n+1}.$ Note that using the given condition, we have \[\frac{x_0}{x_0+1}\geq \frac{1}{x_1+1}+\frac 1{x_2+1}+\cdots+\frac 1{x_n+1}\geq n\sqrt[n]{\frac 1{(x_1+1)(x_2+1)\cdots (x_n+1)}}.\]Getting its analogues and multiplying all the relations together, we get \[x_0x_1x_2\cdots x_n\geq n^{n+1}.\]Equality occurs iff $x_i=n\forall i=0,1,\cdots n.\Box$ This. Was. Beautiful.

Let $x_i = \tan(a_i - \pi/4)$ for each $i$. Then it suffices to show that $$\prod \left(\frac{x_i+1}{1-x_i}\right) \geq n^{n+1} \iff \sum \log\left(\frac{x_i+1}{1-x_i}\right) \geq (n+1) \log n.$$For $y_i = \frac 12(1-x_i)$, the inequality becomes $$\sum \log\left(\frac 1{y_i} - 1\right) \geq (n+1)\log n,$$which is immediate by Jensen. Remark. Interestingly enough, direct Jensen on the first display doesn't quite work, which motivates the substitution.

Define $t_i = \tan(a_i) \ge 0$. Then \[ \sum_{i=0}^n \frac{t_i - 1}{t_i + 1} \ge n - 1 \]\[ 1 \ge \sum_{i=0}^n \frac{1}{t_i + 1} \]By GM-HM it follows that \[ \sqrt[n+1]{(t_0 + 1)\dots (t_n + 1)} \ge \frac{n + 1}{\frac{1}{t_0 + 1} + \dots + \frac{1}{t_n + 1}} \]and thus \[ (t_0 + 1)(t_1 + 1)\dots (t_n + 1) \ge (n+1)^{n+1}. \]Smoothing together $t_i$ and $t_j$ while preserving $(t_i + 1)(t_j + 1)$ decreases $t_it_j$, thus $t_0 \dots t_n$ is minimized when $t_0 = t_1 = \dots = t_n = n$.