Another way to derive it: Try to find the steepest line through the origin that touches $(1+x)^k$, i.e. the largest real number $l$ such that $f(x)=(1+x)^k-lx\ge0$ for all positive $x$. Differentiating yields $f'(x)=k(1+x)^{k-1}-l$ and so we have $f'(x)=0$ for $x_0=\left(\frac{l}{k}\right)^{\frac{1}{k-1}}$. Hence we need $f(x_0)= 0$ to get the optimal $l$. This immediately gives $l=\frac{k^k}{(k-1)^{k-1}}$ and we are done.

Actually, If f(x)=((1+a_n)^n)/a_n then f´(x)=((n-1)(1+a_n)^(n-1))(n(a_n)-1), so the minimum point of f is when x=1/n-1 This proves everyhing, since this yields LHS is at least (a_2)(a_3)...(a_n)(n^n), with equality iff a_n=1/(n-1), so it is impossible for (a_2)(a_3)...(a_n)=1

Does anybody else find it strange that this was chosen as problem # 2?

totally strange, i remember that last year we also got the wrong problems in the beginning.

LMat wrote: Does anybody else find it strange that this was chosen as problem # 2? It looks like a very good choice to me: 1) The students from the top-10 nations will solve it by routine methods; so they will have enough time to work on the more time-consuming problem #3. 2) The students from the medium nations will solve it, but need more time; some will even get stuck with it. 3) The students of the weaker nations do not have the standard tools available. For them it will take a lot of insight to find a solution through Cauchy-Schwarz or through some mean inequality.

This idea of splitting up a constant in such a way is hardly original. Indeed, the main inequality used in Potla's solution $(a_k+1)^k\geq\frac{k^k}{(k-1)^{k-1}}\cdot a_k$ also appears, for example, in lorincz's solution to this Russia 2007 problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=912962

Learner, do you mean too easy?

Does anyone have an idea to solve this problem using Jensen inequality... (after logarithming both of the sides)?

Can somebody hide the solutions this topic because you cannot look at the problem without seeing the solution

Since $a_2 a_3 .... a_n=1$, then set $a_2 =\frac{x_2}{x_3}, a_3=\frac{x_3}{x_4}, ..., a_n=\frac{x_n}{x_2}$. We have to prove that \[ (x_2 +x_3)^2 (x_3 +x_4)^3 ... (x_n +x_2)^n > n^n x_3 ^2 x_4 ^3 ... x_{n}^{n-1} x_2 ^n \] For each $k$ by AM-GM we have that $(x_k +x_{k+1})^{k} =\left(x_k +(k-1)\frac{x_{k+1}}{k-1}\right)^k \geqslant k^k x_k \frac{x_{k+1}^{k-1}}{(k-1)^{k-1}}$. After multiplying all inequalities for $k=2$ to $k=n$, we obtain the desired.

Anzoteh wrote: Actually, If f(x)=((1+a_n)^n)/a_n then f´(x)=((n-1)(1+a_n)^(n-1))(n(a_n)-1), so the minimum point of f is when x=1/n-1 This proves everyhing, since this yields LHS is at least (a_2)(a_3)...(a_n)(n^n), with equality iff a_n=1/(n-1), so it is impossible for (a_2)(a_3)...(a_n)=1 Since the above is hardly understandable, but worth taking a second look, I took the liberty to put it in a readable form. Actually, if we consider the functions $f_k \colon\mathbb{R}_+^* \to \mathbb{R}_+^*$ given by $f_k(x)= \dfrac {(1+x)^k} {x}$, then $f_k'(x)=\dfrac {(1+x)^{k-1}((k-1)x-1)} {x^2}$, so $f_k$ reaches its minimum at $x_k=\dfrac {1} {k-1}$. This proves everything, since $f_k(x_k) = \dfrac {k^k} {(k-1)^{k-1}}$ yields (by telescoping) that LHS is at least $n^n a_2a_3\cdots a_n = n^n$, with equality if and only if $a_k=\dfrac {1} {k-1}$ for each $k$, when it is impossible to have $a_2a_3\cdots a_n=1$.

Unbelievably trivial for IMO 2. This tyoe of inequality should be known to everyone who has ever seen aplication of Am-Gm. My solution is the same as Potla's so I don't want to repost it.

This problem doesn't fit to the IMO. It's too tricky and if you guess this trick the problem becomes obvious. So if you're lucky, you will solve it but if you aren't this problem becomes a ... [mod: no need for bad language here]. A problem like that was on the regional round in Dushanbe. With best regards from Kirovograd

LMat wrote: Does anybody else find it strange that this was chosen as problem # 2? actually, i found it strange that it was selected generally. it was an one-minute problem.

enndb0x wrote: Since $a_2 a_3 .... a_n=1$, then set $a_2 =\frac{x_2}{x_3}, a_3=\frac{x_3}{x_4}, ..., a_n=\frac{x_n}{x_2}$. We have to prove that \[ (x_2 +x_3)^2 (x_3 +x_4)^3 ... (x_n +x_2)^n > n^n x_3 ^2 x_4 ^3 ... x_{n}^{n-1} x_2 ^n \] For each $k$ by AM-GM we have that $(x_k +x_{k+1})^{k} =\left(x_k +(k-1)\frac{x_{k+1}}{k-1}\right)^k \geqslant k^k x_k \frac{x_{k+1}^{k-1}}{(k-1)^{k-1}}$. After multiplying all inequalities for $k=2$ to $k=n$, we obtain the desired. Cool solution! Congratulations! So useful substitution) Exactly this great subtitution solved the problem.

Potla wrote: Equality holds iff $a_k=\frac 1{k-1}$ for all $k,$ which is not possible. $\Box$ Uhhh...for $n=2$, $a_2=1$, we get $(a_2+1)^2=n^n=2^2=4$ and equality does hold, so the assertion we are asked to prove in the original problem statement is, strictly speaking, not always true. For $n>2$ the equality condition would imply $a_{2}\cdot a_{3}\cdot\dots\cdot a_{n}<1$, so I'd agree equality cannot hold for $n>2$. Or am I missing something?

test20 wrote: LMat wrote: Does anybody else find it strange that this was chosen as problem # 2? It looks like a very good choice to me: 1) The students from the top-10 nations will solve it by routine methods; so they will have enough time to work on the more time-consuming problem #3. 2) The students from the medium nations will solve it, but need more time; some will even get stuck with it. 3) The students of the weaker nations do not have the standard tools available. For them it will take a lot of insight to find a solution through Cauchy-Schwarz or through some mean inequality. Exactly for those reasons I think it is a poor choice. I thought IMO problems were supposed to diminish the difference caused by unequal amount of training provided in different countries. Problems which are hard to train for are the only ones which really test your problem-solving ability.

In fact this particular posting of Problem 2 is just one of the many that almost simultaneously appeared, but were removed. Unfortunately, the surviving one (namely, this) misses to state that $n\geq 3$ [mod: now edited!]

We can see that by AM-GM \[\frac{(n-1) \cdot \frac{1}{n-1}+\left(\frac{a_n}{n-1}\right)}{n} \geq \sqrt[n]{\frac{a_n}{(n-1)^{n-1}}} \Rightarrow (1+a_n)^n \geq \frac{n^na_n}{(n-1)^{n-1}}\]If we combine this inequality for all $n$ we get the result. We cannot have the equality case for this because we would need $a_n = \frac{1}{n-1}$ which is obviously impossible since their product is $1$. $\blacksquare$

Note that, by Weighted AM-GM, \begin{align*} \left(1+a_n\right)^n = \left(\dfrac{1}{n - 1} + \dfrac{1}{n - 1} + \cdots + \dfrac{1}{n - 1} + a_n\right)^n \geq \left(\dfrac{n^n}{(n-1)^{n - 1}}\right) \cdot a_n. \end{align*}Therefore, \begin{align*} \prod_{k=2}^{n} (1+a_k)^k \geq \prod_{k=2}^{n} \left(\dfrac{n^n}{(n-1)^{n - 1}}\right) \cdot a_n = n^n. \end{align*}Equality holds if $a_k = \tfrac{1}{k-1}$ for all $1 \leq k \leq n$. Clearly, this cannot happen if $a_1a_2 \cdots a_n = 1$.

Pretty much the same as all of the other solutions: $\left(1+a_n\right)^n = \left(\dfrac{1}{n - 1} + \dfrac{1}{n - 1} + \cdots + \dfrac{1}{n - 1} + a_n\right)^n \geq \left(\dfrac{a_n}{(n-1)^{n - 1}}\right) \cdot n^n.$ Then taking the product of all the terms, most things cancel and we are left with $n^n.$ The inequality is strict as the am-gm equality condition is NOT satisfied, as if $a_k=\frac{1}{k-1}$ for all k, then $a_2a_3\dots a_n \neq 1.$ Q.E.D.

By AM-GM, we know that \begin{align*} 1+a_2 &\ge 2\sqrt{a_2} \\ \frac{1}{2}+\frac{1}{2}+a_3 &\ge 3\sqrt[3]{\left(\frac{1}{2}\right)^2a_3} \\ \frac{1}{3}+\frac{1}{3}+\frac{1}{3}+a_4 &\ge 4\sqrt[4]{\left(\frac{1}{3}\right)^3a_4} \\ &\; \; \vdots \\ \frac{1}{n-1}+\cdots + \frac{1}{n-1} + a_n &\ge n\sqrt[n]{\left(\frac{1}{n-1}\right)^{n-1}a_n} \end{align*}Taking the $i$th inequality from the top to the $i+1$th power (The first one is square, then cubed, etc.), and then multiplying them gives \[(1+a_2)^2 (1+a_3)^3 \dots (1+a_n)^n \ge n^n\]Equality occurs when the individual equalities for each inequality occurs. Equality for AM-GM occurs when all the terms are equal. This is when \[a_i = \frac{1}{i-1}\]However, in this case, the product isn't $1$, so equality can't actually occur, giving the desired inequality.

The key idea is that $$(1+a_i)^i = ((i-1) \cdot \frac{1}{i-1} + a_i)^i$$$$\ge i^i((\frac{1}{i-1})^{i-1} \cdot a_i)$$$$ = \frac{i^i}{(i-1)^{i-1}}a_i.$$Now, $$(1+a_2)^2 \cdots (1+a_n)^n$$$$\ge (\frac{2^2}{1^1} \cdot \frac{3^3}{2^2} \cdots \frac{n^n}{(n-1)^{n-1}}) \cdot (a_2 \cdots a_n)$$$$ = n^n.$$For equality to hold, we require $$a_i = \frac{1}{i-1}\forall i;$$but then $$a_2 \cdots a_n = \frac{1}{2 \cdot 3 \cdots n-1} = \frac{1}{(n-1)!},$$which is never possible. Therefore, equality never holds, and \[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]$\square$

by using the given condition and weighted am-gm we get $$(1+a_i)^i = (\frac{(i-1)}{(i-1)} + a_i)^i \geq \frac{i^ia_i}{(i-1)^{(i-1)}} $$ Now the product nicely cancels out to give the desired result $$(1+a_2)^2(1+a_3)^3\cdots (1+a_n)^n \geq \frac{2^2a_2}{1^1}\frac{3^3a_3}{2^2} \cdots \frac{n^na_n}{{n-1}^{n-1}} = a_2a_3\cdots a_n \cdot n^n = n^n$$Now here equality occurs when $a_i = \frac{1}{i-1}$ for all $i$, which gives $$a_2a_3\cdots a_n \neq 1 $$for $n \geq 3$ which is not true thus equality cannot hold

Seems more like 5M. Write it as $$(1+a_2)^2 \geq (2\sqrt{a_2})^2 = 4a_2$$$$(1+a_3)^3 = (\frac12 + \frac12 + a_3)^3 \geq \frac{27}{4} a^3$$or more generally $$(1+a_k)^k = \left( \underbrace{\frac{1}{k-1} + \frac{1}{k-1} + \dots + \frac{1}{k-1}}_{k-1} + a_k \right)^k \geq \left( k \sqrt[k]{\left( \frac{1}{k-1} \right)^{k-1} a_k} \right) ^k = \frac{k^k}{(k-1)^{k-1}}a_k$$Now telescoping gives the desired result, because all the $a_2, a_3, \dots a_n$ cancel out from the condition. Also equality cannot hold because for equality in the AM-GM's we need $a_k = \frac{1}{k-1}, k \in \{2, 3, \dots n\}$ which does not satisfy our condition. Done

Note that $(1+a_2)^2 \geq 4{a_2}$ due to $AM-GM$. Similarly, $(\frac {1}{2} +\frac {1}{2}+ a_3)^3 \geq \frac{27a_3}{4}$ Generalising, we get $$ (1+a_n)^n \geq \frac{n^n a_n}{(n-1)^{n-1}}$$. Multiplying we get the result since the terms of the sequence have product $1$. Note that equality would hold if $a_n = \frac{1}{n-1}$ but in that case the product isn't $1$ so the inequality must be strict.

Note that $$1+a_k$$$$=\frac{1}{k-1}+\frac{1}{k-1}+\dots+\frac{1}{k-1}+a_k$$$$\ge k\sqrt[k]{\frac{a_k}{(k-1)^{k-1}}}$$by AM-GM. Therefore, $$(1+a_k)^k\ge k^k(\frac{a_k}{(k-1)^{k-1}}).$$Multiplying these expressions together, we know that $$(1+a_2)^2(1+a_3)^3\dots(1+a_n)^n\ge n^n(a_2a_3\dots a_n)=n^n.$$Equality is impossible as that would imply $$\frac{1}{k-1}=a_k$$for every $k$, which is impossible.

OG! Observe that $a_k+1 = a_k+ \frac{1}{k-1}+\frac{1}{k-1}+\frac{1}{k-1} + \cdots \frac{1}{k-1}$(here there are $k$ summands)$ \geq k (\frac{a^k}{(k-1)^{k-1}})^{\frac{1}{k}}$ Thus $(a_k+1)^k \geq (a_k)(\frac{k^k}{(k-1)^{k-1}})$ Taking the product from $k=2$ till $k=n$, we get the desired result.

Note that by AM-GM, $$\left( \frac{1+a_k}{k} \right)^k \geq \frac{1}{(k-1)^{k-1}}a_k \implies (1+a_k)^k \geq \frac{k^k}{(k-1)^{k-1}}.$$Thus multiplying the equations yields $$(1+a_2)^2(1+a_3)^3 (\cdots)(1+a_n)^n \geq n^n$$as the product telescopes. Note that equality holds when $a_k=\frac{1}{k-1}$ for each $k,$ but this is impossible as then $a_2a_3 \cdots a_n < 1.$ Therefore, our inequality is strict so we have shown the desired result. QED

Posting for storage. Consider a single bracketed term. $$(1+a_n)^n$$In such a term, after expanding and counting multiple like terms separately we will have $2^n$ terms. Then applying AM-GM we see that, \begin{align*} (1+a_n)^n &= a_n^n+na_n^{n-1}+\dots + 1\\ &\geq 2^n(\sqrt[2^n]{a_n^{k}})\\ &\geq 2^n(a_n^{\frac{k}{2^n}})\\ & > 2^na_n \end{align*}Here, notice that we obtain the fourth line from $k> 2^n$ ($k=2^n$ when $n=2$) since we are multiplying the coefficient of each term with the power to obtain $k$, but $2^n$ is simply the sum of the coefficients. Then, notice that the $LHS$ of the required inequality, $S$ \begin{align*} S &> 2^2\cdot 2^3 \dots 2^n a_2a_3\cdots a_n\\ &= 2^{\frac{n^2+n-2}{2}} \end{align*}This reduces the desired inequality to $$2^{\frac{n^2+n-2}{2}} \geq n^n$$ Claim : $2^{\frac{n^2+n-2}{2}} \geq n^n$ for all $n \geq 2$ Proof : Now, notice that \begin{align*} &2^{\frac{n^2+n-2}{2}}\geq n^n\\ &\Longleftrightarrow 2^{\frac{n^2+n-2}{2n}}\geq n \end{align*}This inequality is obviously true for $n=2$ (This is when we have equality). Then, if we have $$2^{\frac{n^2+3n}{2n+2}-\frac{n^2+n-2}{2n}}\geq 1$$for all $n$, then the required inequality (second line) will always be true. It is easy to verify this (the LHS is much larger than 1 when $n$ gets larger). Now, notice that this means, \begin{align*} S &> 2^2\cdot 2^3 \dots 2^n a_2a_3\cdots a_n\\ &= 2^{\frac{n^2+n-2}{2}}\\ &\geq n^n \end{align*}which was the required inequality.

cursed_tangent1434 wrote: Posting for storage. Consider a single bracketed term. $$(1+a_n)^n$$In such a term, after expanding and counting multiple like terms separately we will have $2^n$ terms. Then applying AM-GM we see that, \begin{align*} (1+a_n)^n &= a_n^n+na_n^{n-1}+\dots + 1\\ &\geq 2^n(\sqrt[2^n]{a_n^{k}})\\ &\geq 2^n(a_n^{\frac{k}{2^n}})\\ & > 2^na_n \end{align*}Here, notice that we obtain the fourth line from $k> 2^n$ ($k=2^n$ when $n=2$) since we are multiplying the coefficient of each term with the power to obtain $k$, but $2^n$ is simply the sum of the coefficients. Then, notice that the $LHS$ of the required inequality, $S$ \begin{align*} S &> 2^2\cdot 2^3 \dots 2^n a_2a_3\cdots a_n\\ &= 2^{\frac{n^2+n-2}{2}} \end{align*}This reduces the desired inequality to $$2^{\frac{n^2+n-2}{2}} \geq n^n$$ Claim : $2^{\frac{n^2+n-2}{2}} \geq n^n$ for all $n \geq 2$ Proof : Now, notice that \begin{align*} &2^{\frac{n^2+n-2}{2}}\geq n^n\\ &\Longleftrightarrow 2^{\frac{n^2+n-2}{2n}}\geq n \end{align*}This inequality is obviously true for $n=2$ (This is when we have equality). Then, if we have $$2^{\frac{n^2+3n}{2n+2}-\frac{n^2+n-2}{2n}}\geq 1$$for all $n$, then the required inequality (second line) will always be true. It is easy to verify this (the LHS is much larger than 1 when $n$ gets larger). Now, notice that this means, \begin{align*} S &> 2^2\cdot 2^3 \dots 2^n a_2a_3\cdots a_n\\ &= 2^{\frac{n^2+n-2}{2}}\\ &\geq n^n \end{align*}which was the required inequality. splendid solution!

Tried a $a_i = \frac{b_i}{b_{i+1}}$ substitution + induction that went nowhere :C I cannot do inequalities

XUH!!!!!!! THIS IS BANNED \[\prod_{k=2}^{n}(1+a_k)^k=\prod_{k=2}^{n}\left((k-1)\cdot\frac{1}{k-1}+a_k\right)^k \ge\prod_{k=2}^{n}\frac{k^k a_k}{(k-1)^{k-1}}=k^k.\]BUT:: EQUALITY CASE DOESNT WORK!!!!!!!!!!!! BECAUSE $a_k=\frac{1}{k-1}$ DON'T MULTIPLY TO ONE .