AoPS Academy
be an acute triangle with 
. The angle bisector of 
intersects 
at 
, and the perpendicular to 
at intersects 
and 
at 
respectively. Suppose 
and 
intersect again at 
, prove that 
is the angle bisector of 
.
and 
Prove that
Let and 
Prove that
![\[ \begin{tabular}{c c c c}
{} & \textbf{Adams} & \textbf{Baker} & \textbf{Adams and Baker} \\
\textbf{Boys:} & 71 & 81 & 79 \\
\textbf{Girls:} & 76 & 90 & ? \\
\textbf{Boys and Girls:} & 74 & 84 & \\
\end{tabular}
\]](http://latex.artofproblemsolving.com/8/5/c/85c75770c3f90af49df0db600dead2a29e6674b3.png)
. The inequality becomes 
, where 
are positive and 
. This is a mere application of AM-GM to the denominators 
. We get ![$A_i\ge n\sqrt[n]{\frac{\prod a_k}{a_i}}$](http://latex.artofproblemsolving.com/6/f/5/6f53a9d420b96b8d7d298c3c861f0d4e2d82ebb6.png)
. After multiplying all of these we get the desired result.
such that 
. Cross multiplication and substitution yields that the desired inequality is equivalent to![\[n^{n+1} \le \frac{(1-a_1)(1-a_2) \dots (1-a_{n+1})}{a_1 a_2 \dots a_{n+1}} \quad (*)\]](http://latex.artofproblemsolving.com/8/d/f/8df0524a918d91dd5de085ab7d77bc159bacbe84.png)
![\[n\sqrt[n] {\frac{a_1 a_2 \dots a_{n+1}}{a_i}} \le a_1 + a_2 + \dots + a_{i-1} + a_{i+1} + \dots + a_{n+1} = 1-a_i\]](http://latex.artofproblemsolving.com/1/a/c/1ac4cf8b3d11ef8f829b070fa8ecfd57418bdb57.png)
and dividing by 
yields 
and the proof is complete.
. The algebra of verifying the smoothing is tedious but straight forward.
. It becomes
which is all less than 
.
be positive real numbers such that 
. Prove that![\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]](http://latex.artofproblemsolving.com/6/5/5/65558feba82e32266d3d3cbdfea85e079483403f.png)
as 
(it really is decoration). Then we just want 
. However, notice that ![$1-a_i=a_1+\dots+a_{i-1}+a_{i+1}+\dots+a_n+b \geqslant n\sqrt[n]{a_1\dots a_{i-1}a_{i+1}\dots a_nb}$](http://latex.artofproblemsolving.com/e/3/1/e31182ce7a05ae323004f7ad40b8abda6ef457ec.png)
and multiplying all of them gives the result. 
have fixed sum 
. Note![\[\frac{ab}{(1-a)(1-b)}\le t\iff ab\le t-tc+tab\iff ab(1-t)\le t(1-c)\]](http://latex.artofproblemsolving.com/7/9/7/797c3c911e24373f699d405ddc46376038b325c5.png)
where 
is the maximum of the expression.
and so equality would have to be achieved at 
. Thus by a smoothing argument each is equal. Let 
so the desired is![\[\frac{1}{n^{n+1}}\ge \frac{\frac{c^n}{n^n}[1-c]}{c\cdot (1-\frac cn)^n}=\frac{c^{n-1}[1-c]}{(n-c)^n}\iff (n-c)^n\ge n^{n+1}c^{n-1}[1-c].\]](http://latex.artofproblemsolving.com/a/8/d/a8dfd7597dc371de3737181110df456b78c25661.png)
The result is obvious at 
so we can assume 
.
so we need to check![\[\left(n^2+d\right)^n \ge n^{n+1}\left(n-d\right)^{n-1}\left(1+d\right)\]](http://latex.artofproblemsolving.com/a/8/d/a8da6614035c7634c6c5afdb797cdd80632e5558.png)
and have 
. Since the result is true at the extrema, it suffices to check cases where the derivatives of both sides are equal. In these cases,![\[n(n^2+d)^{n-1}=-(n-1)n^{n+1}(n-d)^{n-2}\cdot (1+d)+n^{n+1}(n-d)^{n-1}.\]](http://latex.artofproblemsolving.com/e/6/c/e6c2dfca54371b44bad1398f2eff7933fcdfd128.png)
Equivalently,![\[(n^2+d)^{n-1}=n^n(n-d)^{n-2}\cdot [(n-d)-(n-1)(1+d)]=n^n(n-d)^{n-2}\cdot [1-nd].\]](http://latex.artofproblemsolving.com/9/2/a/92a8a6e5fa78e646fc66f4c924af8a727bc2d87f.png)
Equality can only hold at 
because 
increases the left side and decreases the right side whereas 
increases the right side and decreases the left side. But at the result is obvious, so we are done.
. Thus, our expression becomes![\[\prod_{i=1}^{n+1} \frac{a_i}{1-a_i} = e^{\sum \ln(\frac{a_i}{1-a_i})}\]](http://latex.artofproblemsolving.com/9/0/5/90541a29af135fa6d5e2529c4865a7c81c5832d3.png)
Note that 
has a second derivative of 
and is thus convex over 
, thus, by jensens![\[\sum \ln(\frac{a_i}{1-a_i}) \geq (n+1)\cdot \ln(\frac{1}{n})\]](http://latex.artofproblemsolving.com/6/3/3/6337df2ca899c5514bf8e40988dd55c0562abd4e.png)
Thus, ![\[\prod_{i=1}^{n+1} \frac{a_i}{1-a_i}\geq e^{(n+1)\cdot \ln(\frac{1}{n})} = \left(\frac{1}{n}\right)^{n+1}\]](http://latex.artofproblemsolving.com/e/6/a/e6a273e89e1d79bcb59d5752df1521a788298388.png)
and we are done .![\[\frac{2x-1}{(x-1)^2x^2}\]](http://latex.artofproblemsolving.com/1/b/6/1b61078a4304ffaf2a444a48b6511b681950fd3c.png)
New Correct Solution.![\[\prod_{i=1}^{n+1} \sum_{j\neq i} a_i \leq \prod_{i=1}^{n+1} n\cdot \sqrt[n]{\prod_{j\neq i}a_i} \leq n^{n+1} \cdot \prod a_i\]](http://latex.artofproblemsolving.com/f/4/a/f4a40b6a93ab14b4d13601e668dd5834b9fa4623.png)
But wait! this is the denominator of our new expression. The given expression becomes![\[\frac{\prod a_i}{\prod (1-a_i)}=\frac{\prod a_i}{\prod (\sum_{j\neq i} a_j)}\leq \frac{\prod_{i=1}^{n+1}}{n^{n+1}\prod_{i=1}^{n+1}}=\frac{1}{n^{n+1}}\]](http://latex.artofproblemsolving.com/c/b/8/cb8de4a63db7c71c27ab3b606dfc76a69931e9f8.png)
and we are done.
. Then we must have 
, and we want to prove that 
Now notice that ![$$1-a_i=(a_1+\ldots+a_{n+1})-a_i\ge n\sqrt[n]{\frac{a_1\ldots a_{n+1}}{a_i}}.$$](http://latex.artofproblemsolving.com/b/4/f/b4fb3bc8026a5840b40a409ba2461a28f7a02a8f.png)
Applying this to all the terms in the denominator gives us ![\begin{align*} \frac{a_1\ldots a_{n+1}}{(1-a_1)\ldots(1-a_{n+1})} & \le \frac{a_1\ldots a_{n+1}}{n^{n+1}\sqrt[n]{\frac{a_1\ldots a_{n+1}}{a_1}}\ldots \sqrt[n]{\frac{a_1\ldots a_{n+1}}{a_{n+1}}}} \\ &= \frac{a_1\ldots a_{n+1}}{n^{n+1}\sqrt[n]{a_1^n\ldots a_{n+1}^n}} \\ &= \frac{a_1\ldots a_{n+1}}{n^{n+1}(a_1\ldots a_{n+1})} \\ &= \frac{1}{n^{n+1}}.\end{align*}](http://latex.artofproblemsolving.com/9/b/9/9b981faf7129694475c7e3b4d8aab7c5d9645392.png)
The proof is complete.
, we have 
Then I find 
so Jensen doesn't work, but I see that there is exactly one inflection point in 
, so 
EV is applicable (you should not need EV on an A1)... You can see how this approach of trying fancy things first fails miserably.
![$$\sqrt[n]{\left(\frac{1}{a_1}-1\right) \left(\frac{1}{a_2}-1\right)\ldots \left(\frac{1}{a_n}-1\right)}\geq \frac{n}{\frac{a_1}{1-a_1}+\frac{a_2}{1-a_2}+\ldots+\frac{a_n}{1-a_n}}.$$](http://latex.artofproblemsolving.com/f/e/6/fe6568766e06a91a573901e9150f53610c7cefe1.png)
Now it is suffices to show that
Therefore it is suffices to show that
Let 
.![\[ \left(t-\frac{1}{n}\right)^{n} \geq n\left(t-1\right).\]](http://latex.artofproblemsolving.com/f/2/8/f28facdbeaeb01f005056b3cbe54138cded2e415.png)
Define 
, such that 
.
.
and 
.
and thus the minimum of the function is
Hence, we have proven the inequality ![\[ \left(t-\frac{1}{n}\right)^{n} \geq n\left(t-1\right),\]](http://latex.artofproblemsolving.com/e/e/5/ee57607e188f0a8b871a41a1facc3d27ea2ca7cc.png)
where the equality holds iff 
. We conclude that that the equality in our original inequality holds iff 
. We are done.
for positive reals 
with 
that sum to 
(in the problem, is replaced with 
).![$$\frac{1}{a_1}-1=\frac{a_2+\dots+a_{n+1}}{a_1}\geq \frac{n\sqrt[n]{a_2a_3\dots a_{n+1}}}{a_1}$$](http://latex.artofproblemsolving.com/f/5/9/f590f3ffb0abf9e03425abd3fbba65e1e989f4dd.png)
and multiplying together gives the desired result. Equality holds when 
.
, let 
, and let 
. Note
and all 
are positive.
,![$$\sum_{j\neq k}b_k\geq n\sqrt[n]{\prod_{j\neq k}b_k}$$](http://latex.artofproblemsolving.com/a/d/a/ada1a074ae95dadd3414764b40df1cf237bbdbee.png)
So,![$$\prod_{k=1}^{n+1} \sum_{j\neq k}b_j\geq \prod_{k=1}^{n+1} n\sqrt[n]{\prod_{j\neq k}b_j}$$](http://latex.artofproblemsolving.com/8/e/6/8e6f9d8ac4835e6c773a7ab63e698b43c8326df7.png)
![$$\prod_{k=1}^{n+1} \sum_{j\neq k}b_j\geq n^{n+1}\prod_{k=1}^{n+1} \sqrt[n]{\prod_{j\neq k}b_j}$$](http://latex.artofproblemsolving.com/0/7/9/079facfcda7b4f923c4dbaf5022656c994af6e20.png)
![$$\frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}$$](http://latex.artofproblemsolving.com/b/e/9/be9059bcffa18ab7107879fe7a9beeb742ed6c9f.png)
case and define 
. Then, we have 
, and we want to show that ![\[ \prod\limits_{i=0}^{n} \frac{a_i}{1-a_i} \leq \frac{1}{n^{n+1}} \]](http://latex.artofproblemsolving.com/a/f/a/afab9cf58a68b19c39d39c5e13be2d2337cb06b2.png)
and notice that equality holds when 
. Now, we smooth. Say that there is some set of values 
such that the value of the product is maximal, and assume for the sake of contradiction that there exist some 
. Then, we are done if ![\[ \frac{a_i}{1-a_i}\frac{a_j}{1-a_j} \leq \left(\frac{a_i+a_j}{2-a_i-a_j}\right)^2 \]](http://latex.artofproblemsolving.com/b/c/e/bce9cd3caafdb0aeade6495eba5331e0a20be0ad.png)
as that would imply that there was some larger value of the product by taking the values ![\[ \left(a_0,a_1,\ldots,a_{i-1},\frac{a_i+a_j}{2},a_{i+1},\ldots,a_{j-1},\frac{a_i+a_j}{2},a_{j+1},\ldots,a_n\right) \]](http://latex.artofproblemsolving.com/e/2/e/e2ecc5d9f2f264a62c9ea07ccc7637aa1f745653.png)
Now, we prove that this set is actually larger. We use the subsitution 
and 
. Then, we can rewrite the equation we want to prove as ![\[ \frac{a_i}{1-a_i}\frac{a_j}{1-a_j} \leq \left(\frac{a_i+a_j}{2-a_i-a_j}\right)^2 \Rightarrow \frac{u^2 - v^2}{(1-u)^2 - v^2} < \frac{u^2}{(1-u)^2} \]](http://latex.artofproblemsolving.com/b/7/9/b79bc00de956bd357f4e54a32017fd7fa25b2675.png)
after some algebra, which further rearranges into ![\[ \frac{u^2 - (1-u)^2}{(1-u)^2 - v^2} < \frac{u^2 - (1-u)^2}{(1-u)^2} \]](http://latex.artofproblemsolving.com/f/2/0/f20012b512f63284d515d9a4eb98824181ba1cd8.png)
which is true because 
.
be the positive real number such that 
The desired inequality becomes ![\[ (na_{1}) (na_{2}) \cdots (na_{n}) (na_{n+1}) \le ( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})(1-a_{n+1}) \]](http://latex.artofproblemsolving.com/8/a/9/8a92688a8f523074c14d5a27e92eaf7db6c8a192.png)
By AM-GM we have ![\[1-a_{i}=a_1+a_2+\dots+a_{n+1}-a_i\ge n\cdot \sqrt[n]{\frac{a_1a_2\cdots a_{n+1}}{a_i}}\]](http://latex.artofproblemsolving.com/e/5/4/e54ec1dacd62e05b4ecaf2782a774ada82c82c70.png)
Multiplying the analogous gives the result.
. Taking 
of both sides, we can first optimize 
Using log properties, we find that the second derivative of the term we are summing is 
which is zero at 
, so there is exactly one inflection point, meaning that we can use EV to have WLOG 
. Our wanted inequality is now 
By AM-GM, we can instead just show that 
Adding 
to both sides, we want 
We can resort to differentiating the LHS with respect to 
to get 
which is zero when 
. The second derivative is positive for the whole interval, so that root is the global minimum for the interval. Plugging this back in, we want to show that 
We can do this using another direct differentiation with respect to 
. We get 
and we have to check 
It is easy to check that the wanted inequality does hold true for all of these, so we are done.
. Then, the inequality is equivalent to 
However, observe that ![$$1-a_1 = a_2+a_3+\cdots+a_{n+1} \geq n\sqrt[n]{a_2a_3\cdots a_{n+1}},$$](http://latex.artofproblemsolving.com/8/e/1/8e1cd7d48d1b69520caf05b9c69ec689f32aedc0.png)
so multiplying this inequality cyclically in the denominator yields the result.
, thus the inequality is equal to: 
![\begin{align*} a_1+\dots +a_n\ge n\sqrt[n]{\prod_{i=1}^n a_i}\end{align*}](http://latex.artofproblemsolving.com/e/5/d/e5d760b2e443ce5a9968c300ea96048c8b233619.png)
![\begin{align*}a_1+\dots +a_{n-1}+a_{n+1}\ge n\sqrt[n]{a_1\cdots a_{n-1}a_{n+1}} \end{align*}](http://latex.artofproblemsolving.com/0/7/5/075792dd906017c586757745c0f21fcb5ef6dc36.png)
By multiplying the inequalities: 
. QED
. We claim the maximum can be achieved when all the variables are equal. Assume not. WLOG, 
. Let 
. But consider![\[
\frac{\left( \frac{x+y}2 \right)^2}{\left(1 - \frac{x+y}2 \right)^2} - \frac{xy}{(1-x)(1-y)} = \frac{(x-y)^2 (1-x-y)}{(1-x)(1-y)(2-x-y)^2} \ge 0,
\]](http://latex.artofproblemsolving.com/e/1/c/e1c336ec312c78650e6a63a359f57c643405e4f7.png)
so making them equal (with the same sum) results in a product at least as big as the original one. The conclusion follows.
be positive real numbers such that . Prove that 
, it suffices to prove that 
. Note that 1-a_i=the sum of all a_j's from 1 to n excluding a_i. Then this inequality follows immediately from 
, where the greater or equal to is just the AM-GM. Each of these steps are reversible, so we're done.
with 
, now we have the better condition 
, prove 
. We proceed by smoothing, we show replacing any two with their average does not decrease the product. We desire 
, equivalently 
, standard expansion gives the desired as 
, cancellation gives 
. We can write this as 
. We can now write the left hand side as 
, since 
, the bound is now obvious. repeatedly, we can see approaches 
. Assume there exists some tuple 
for which the product is a factor of 
greater than the claimed maximum. After applying the operation some arbitrarily large number of times, we can eventually reach a tuple 
for which each element is a factor 
away from for arbitrarily small , so we can then bound the product for as 
away from desired for arbitrarily small , choosing 
forces a contradiction.
specifies the greatest common divisor of 
and 
which is the same as
However, by AM-GM, we have![$$a_1+a_2+\ldots+a_n\geq n\sqrt[n]{a_1a_2\ldots a_n},$$](http://latex.artofproblemsolving.com/6/8/6/686487873f223705e64961827a354ef9aca65e90.png)
so the denominator is greater than or equal to 
Therefore, the entire fraction is less than or equal to 
.
,
and 
.
Taking the product over 
that is,
and take the reciprocal to get that the original inequality is equivalent to ![\[\left(\frac{1-a_1}{a_1}\right)\cdot \left(\frac{1-a_2}{a_2}\right)\cdots \left(\frac{1-a_{n+1}}{a_{n+1}}\right)\ge n^{n+1}.\]](http://latex.artofproblemsolving.com/0/4/0/0409ed2ae36819f21a2d4406cf9fd07fc7530add.png)
![$1-a_i=\sum_{k=1, k\ne i}^{n+1} a_k\ge n\sqrt[n]{\prod_{k=1, k\ne i}^{n+1} a_k}$](http://latex.artofproblemsolving.com/a/e/d/aed4f7318592ce201e8ae7730916138b46070f0b.png)
.
.
.
.![$$\frac{1}{a_1}-1 = \frac{a_2+a_3 + \cdots + a_{n+1}}{a_1} \ge \frac{n \sqrt[n]{a_2a_3 \cdots a_{n+1}}}{a_1}$$](http://latex.artofproblemsolving.com/2/9/d/29d65bf18b1a6e13fd779c1988201b6498d6a597.png)
and multiplying similar inequalities gives the result.
;![\[\frac{a_1a_2\dots a_{n+1}}{\prod_{\text{cyc}}(a_1+a_2+\dots+a_n)}\le \frac{a_1a_2\dots a_{n+1}}{\prod_{\text{cyc}}(n\sqrt[n]{a_1a_2\dots a_n})}=\frac{1}{n^{n+1}}\]](http://latex.artofproblemsolving.com/3/4/4/3440dca042d428d38188864079d3e54c987321d6.png)
so we are done. 
.
.
,![\begin{align*}
\sqrt[n]{\prod_{i\neq k}a_i} &\leq \dfrac{\sum_{i\neq k}a_i}n\\
\implies \sqrt[n]{\prod_{i\neq k}a_i} &\leq \dfrac{(1-a_k)}n
\end{align*}](http://latex.artofproblemsolving.com/5/a/5/5a525d2fd4ab25ff381db0e162b1c390bef4c378.png)
And, multiplying all of these gives,
Which is what we wanted. 
,
.
which is clearly true by AM-GM on the numerator. 
,![\[
\frac{a_1}{1-a_1} \cdot \frac{a_2}{1-a_2}
\cdots \frac{a_n}{1-a_n} \cdot \frac{1-S}{S}
\leq \frac{1}{n^{n+1}}.
\]](http://latex.artofproblemsolving.com/9/9/c/99c8d48ae91591abe99fab4165456301fa2b1953.png)
![\[
\frac{a_1}{1-a_1} \cdot \frac{a_2}{1-a_2} \cdots \frac{a_n}{1-a_n}
\leq \frac{1}{n^n} \cdot
\bigg(\frac{a_1}{1-a_1} + \frac{a_2}{1-a_2} + \cdots + \frac{a_n}{1-a_n}\bigg),
\]](http://latex.artofproblemsolving.com/b/2/1/b21fd7a42df5dc4f62ff56a8d84709bf6b56a4f0.png)
so we want to prove![\[
\bigg(\frac{a_1}{1-a_1} + \frac{a_2}{1-a_2} + \cdots + \frac{a_n}{1-a_n}\bigg)^n
\cdot \frac{1-S}{S} \leq \frac1n.
\]](http://latex.artofproblemsolving.com/2/8/8/288aee65158049ac96a56c2f5a5c67d9e6d00683.png)
,
so 
.
as 
and 
for 
where 
,
.![\[
\frac{a_1}{1-a_1} + \frac{a_2}{1-a_2} + \cdots + \frac{a_n}{1-a_n}
\leq \frac{b_1}{1-b_1} + \frac{b_2}{1-b_2} + \cdots + \frac{b_n}{1-b_n}.
\]](http://latex.artofproblemsolving.com/2/c/e/2ceaa13efb0706364ae001dbbba4a6cbdda51b1d.png)
![\[
\bigg(\frac{S-(n-1)t}{1-S+(n-1)t} - (n-1) \cdot \frac{t}{1-t}\bigg)^n
\cdot \frac{1-S}{S} \geq \frac1n.
\]](http://latex.artofproblemsolving.com/b/7/8/b7862dc162b45570f664c306e5b1f908a3a9728b.png)
Define![\[
g(t) = (n-1) \cdot \frac{t}{1-t} + \frac{S-(n-1)t}{1-S+(n-1)t}
= (n-1) \cdot \frac{1}{1-t} + \frac{1}{1-S+(n-1)t} - n,
\]](http://latex.artofproblemsolving.com/8/c/4/8c409f5b44a0db693c4120f25143b1243afc1c95.png)
and note that![\[g'(t) = \frac{n-1}{(1-t)^2} - \frac{n-1}{(1-S+(n-1)t)^2}\]](http://latex.artofproblemsolving.com/4/c/a/4cae7e47476ceee65a16c82fe1ffad9f81d1178e.png)
is 
when 
,
is constant.
and 
are impossible since 
and 
.![\[
\bigg(n \cdot \frac{S/n}{1-S/n}\bigg)^n \cdot \frac{1-S}{S}
= \frac{S^{n-1}(1-S)}{(1-S/n)^n} \leq \frac1n \tag{$*$}
\]](http://latex.artofproblemsolving.com/8/7/b/87bc2842390c057b85193be86f1f2e9616c84388.png)
Define![\[
h(S) = \frac{S^{n-1}(1-S)}{(1-S/n)^n},
\qquad h'(S) = \frac{(n-1)S^{n-2}}{(1-S/n)^{n+1}}
\cdot \bigg(1 - \frac{n+1}{n} \cdot x\bigg),
\]](http://latex.artofproblemsolving.com/e/d/2/ed20c331662ab1fbbd347bd0d08e9bc5f8be6765.png)
which is only 
.
)
.
,
Now, note that by AM-GM![\begin{align*}
\frac{a_1a_2\cdots a_{n+1}}{(1-a_1)(1-a_2)\cdots (1-a_{n+1})} &= \prod_{\text{cyc}} \frac{a_1}{1-a_1} \\ &= \prod_{\text{cyc}} \frac{a_1}{a_2+a_3+\cdots + a_{n+1}} \\ &\leq \prod_{\text{cyc}} \frac{a_1}{n\sqrt[n]{a_2a_3\cdots a_{n+1}}} \\ &= \frac{1}{n^{n+1}},
\end{align*}](http://latex.artofproblemsolving.com/b/8/8/b887ea1590ce92790c5c5f8ebcf0a6cd89a1ca34.png)
as desired. 
.
This is easy with AM-GM, as 
to 
,
instead of 
.
.
.
,
,
.
such that 
.![\begin{align*} & \frac{a_2+a_3+\dots+a_{n+1}}n \cdot \frac{a_1+a_3+\dots+a_{n+1}}n \cdot \frac{a_1+a_2+a_4+\dots+a_{n+1}}n \cdots \frac{a_1+a_2+\dots+a_{n-1}+a_{n+1}}n \cdot \frac{a_1+a_2+\dots+a_{n-1}+a_n}n \\ & \qquad \geq \sqrt[n]{a_2a_3 \cdots a_{n+1}}\sqrt[n]{a_1a_3 \cdots a_{n+1}}\sqrt[n]{a_1a_2 a_4 \cdots a_{n+1}}\cdots \sqrt[n]{a_1a_2\cdots a_{n-1}a_{n+1}}\sqrt[n]{a_1a_2 \cdots a_{n-1}a_n}\end{align*}](http://latex.artofproblemsolving.com/9/c/4/9c46472e6ea115eec2aa47b8d821d64a7865dcc5.png)
which is just AM-GM.
Then the inequality is equivalent to 
which is true by AM-GM. QED
is less than 
.
.