Let $S=1+\sum_{j=1}^{n-1} \prod_{i=1}^{n-j} a_i$ and note that by Chebyshev's inequality, we have\begin{align*} nS=\left(\sum_{j=1}^{n-1} \prod_{i=1}^{n-j} a_i+1\right) \cdot \left(\sum_{i=1}^{n} (a_{n-i+1}-1)\right) \le n\cdot \left(\left(\sum_{j=1}^{n-1} \left(\prod_{i=1}^{n-j} a_i\right)\cdot (a_{n-j+1}-1)\right)+1 \cdot (a_1-1) \right)=n(a_1\dots a_n-1) \end{align*}proving the claim. $\blacksquare$

Hmm had a solution similar to drunken master but anyways...

Superguy wrote: Isnt it a NT ?:maybe Its is obvious to note that $a_{n}\geq2$ for all $n$ Then applying induction on $n$ we can get the result. How canu conclude an>=2 for all n?? Its actually wrong

My solution in exam. Observe that RHS is an increasing function in $a_n$ and LHS is independent of $a_n$. Clearly $na_n \geq \sum a_i =2n \implies a_n \geq 2$. Thus the inequality is sharpest at $a_n=2$ and thus we can keep reducing the number of variables by 1 upto $n=2$. The case $n=2$ is equivalent to $(a_1-1)(a_2-2) \leq 0$ which is clear .

hellomath010118 wrote: My solution in exam. Observe that RHS is an increasing function in $a_n$ and LHS is independent of $a_n$. Clearly $na_n \geq \sum a_i =2n \implies a_n \geq 2$. Thus the inequality is sharpest at $a_n=2$ and we are done by induction. The case $n=2$ is equivalent to $(a_1-1)(a_2-2) \leq 0$ which is clear . After clearing $a_n$ you will be left with ineq with constraint $a_1+...a_{n-1} \le 2n-2$ now you cannot assume $a_{n-1}\ge 2$ and hence cannot proceed with the induction

kapilpavase wrote: hellomath010118 wrote: My solution in exam. Observe that RHS is an increasing function in $a_n$ and LHS is independent of $a_n$. Clearly $na_n \geq \sum a_i =2n \implies a_n \geq 2$. Thus the inequality is sharpest at $a_n=2$ and we are done by induction. The case $n=2$ is equivalent to $(a_1-1)(a_2-2) \leq 0$ which is clear . After clearing $a_n$ you will be left with ineq with constraint $a_1+...a_{n-1} \le 2n-2$ now you cannot assume $a_{n-1}\ge 2$ and hence cannot proceed with the induction Kapil the pro strikes again, first Jupiter, now hello math. ... My solution : Chebyshev Inequality on exactly what AM has said

I have assumed $a_n=2$

See what happens is assuming $a_n=2$ applies constraints on $a_{n-1}$ and hence on LHS. It's like bigger value of $a_n$ could lead to bigger value of $a_{n-1}$ and hence bigger value of LHS So you can't assume $a_n=2$

Correct me if I am wrong, but I think the inequality can be easily solved using induction on $n$ and applying the induction hypothesis to $a_1,a_2, \dots, a_{n-2}, a_n +a_{n-1} -2$. I will post the solution once I am on a better device.

Supercali wrote: Correct me if I am wrong, but I think the inequality can be easily solved using induction on $n$ and applying the induction hypothesis to $a_1,a_2, \dots, a_{n-2}, a_n +a_{n-1} -2$. I will post the solution once I am on a better device. Ok let's say you want to prove $abc+ab+a+2\le abcd$. Applying induction hypothesis to $a,b,c+d-2$ we get $ab+a+2\le ab(c+d-2)$ i dont see how this can help

Then you have to prove $2c+d \leq cd+2$ which is just $$(c-1)(d-2) \geq 0$$

Induction (not sure if its correct)- easy to check for $n=2$. Now let $a_n = 2+x, x \geq 0$, we have $a_1 +a_2 +...+(a_{n-1}+x)=2n-2$, so $$2+a_1 +a_1 a_2 +...+a_1 a_2 ... a_{n-2} \leq a_1 a_2 ... a_{n-2} (a_{n-1} +x)$$Now, $$LHS=a_1 a_2 ... a_{n-1} + (2+a_1 +a_1 a_2 +...+a_1 a_2 ... a_{n-2}) \leq a_1 a_2 ... a_{n-2} ( 2a_{n-1} + x ) \leq a_1 a_2 ... a_{n-2} a_{n-1} a_{n}$$as $a_{n-1} a_{n} = 2a_{n-1} + a_{n-1} x \geq 2a_{n-1} + x$

anantmudgal09 wrote: Let $n \geqslant 2$ be an integer and let $1<a_1 \le a_2 \le \dots \le a_n$ be $n$ real numbers such that $a_1+a_2+\dots+a_n=2n$. Prove that$$a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n-2}+\dots+a_1a_2+a_1+2 \leqslant a_1a_2\dots a_n.$$ Proposed by Kapil Pause Needless to say, this is my problem. The solution i submitted is same as one anant has already posted. I am very much eager to see other possible solutions

Supercali wrote: Then you have to prove $2c+d \leq cd+2$ which is just $$(c-1)(d-2) \geq 0$$ Can you elaborate

kapilpavase wrote: See what happens is assuming $a_n=2$ applies constraints on $a_{n-1}$ and hence on LHS. It's like bigger value of $a_n$ could lead to bigger value of $a_{n-1}$ and hence bigger value of LHS So you can't assume $a_n=2$ I don't understand what you mean ... If I assume small $a_n$ then I can choose larger $a_{n-1}$.

kapilpavase wrote: Supercali wrote: Then you have to prove $2c+d \leq cd+2$ which is just $$(c-1)(d-2) \geq 0$$ Can you elaborate Since $d$ is the largest among $a,b,c,d$, we must have $d \geq 2$. Now, applying induction hypothesis on $a,b,c+d-2$ we get $$abc + ab + a + 2 \leq abc + ab(c+d-2) = ab(2c+d-2)$$But, we have $(c-1)(d-2) \geq 0$ which implies $2c+d-2 \leq cd$ and we are done. Basically, Drunken_Master's solution.

Ok....i think yours and drunken master's induction solutions are right ... Nice

hellomath010118 wrote: kapilpavase wrote: See what happens is assuming $a_n=2$ applies constraints on $a_{n-1}$ and hence on LHS. It's like bigger value of $a_n$ could lead to bigger value of $a_{n-1}$ and hence bigger value of LHS So you can't assume $a_n=2$ I don't understand what you mean ... If I assume small $a_n$ then I can choose larger $a_{n-1}$. All you can do is assume $a_n\ge 2$. You cannot forecast that inequality will be tight at $a_n=2$ because a bigger value of $a_n$ might increase LHS (and even if it doesn't you have to prove it..i dont think it's simple) The correct way to induct is as in drunken master's or supercali's solution.

kapilpavase wrote: hellomath010118 wrote: kapilpavase wrote: See what happens is assuming $a_n=2$ applies constraints on $a_{n-1}$ and hence on LHS. It's like bigger value of $a_n$ could lead to bigger value of $a_{n-1}$ and hence bigger value of LHS So you can't assume $a_n=2$ I don't understand what you mean ... If I assume small $a_n$ then I can choose larger $a_{n-1}$. All you can do is assume $a_n\ge 2$. You cannot forecast that inequality will be tight at $a_n=2$ because a bigger value of $a_n$ might increase LHS (and even if it doesn't you have to prove it..i dont think it's simple) The correct way to induct is as in drunken master's or supercali's solution. I am using the wrong vocabulary. I will explain later what I mean. Btw sorry to everyone ....

How many marks can be deducted if you forget to prove the base case n=3 in an induction solution??

I think base case is $n=2$ which requires more or less same idea as the induction step and hence at the most $3$ marks will be deducted. But if your induction proof is totally different I don't know.

This is more or less the solution in #2 , but still posting as it is easier to understand.(I hope !!) $a_1a_2\cdots a_n=a_1a_2\cdots a_{n-1}(a_n-1)+a_1a_2\cdots a_{n-1}$ $=a_1a_2\cdots a_{n-1}(a_n-1)+a_1a_2\cdots a_{n-2}(a_{n-1}-1)+a_1a_2\cdots a_{n-2}$ .. .. $= a_1a_2\cdots a_{n-1}(a_n-1)+a_1a_2\cdots a_{n-2}(a_{n-1}-1)+\cdots +a_1(a_2-1)+1(a_1-1) +1 $ By applying Chebyshev's on the similarly ordered sets ($a_1-1,a_2-1,\cdots,a_n-1$) and ($1,a_1,a_1a_2,\cdots,a_1a_2..a_{n-1}$) we get that $ a_1a_2\cdots a_{n-1}(a_n-1)+a_1a_2\cdots a_{n-2}(a_{n-1}-1)+\cdots +a_1(a_2-1)+1(a_1-1) \geq \frac 1n (1+a_1+a_1a_2+\cdots+a_1,a_2..a_{n-1})(a_1+a_2+\cdots+a_n+-n)$ And hence $a_1a_2..a_n-1\geq 1+a_1+a_1a_2+\cdots+a_1,a_2..a_{n-1} $ Equality iff $a_1=a_2=..=a_n=2$ Hence proved. @ below can you elaborate.

What if we write 1/ai≤1/a1 Then apply the concept of GP.. and then get inequality in a1 only and check the minimum value at a1=a2=...=an=2?

Let P(a_1,a_2,...,a_n)=a_1a_2...a_{n-1}+....a_1+2-a_1a_2...a_n. We note that P is convex wet each a_i. Now it is easy to see that a_i is in the interval (1,n+1) for each a_i. So P attains maximum value when each a_i is either 1 or n+1. Now note that since a_1+a_2+...a_n=2n, only one a_i can be n+1. So maximum is when a_1=a_2=....=a_{n-1}=1 and a_n=n+1. But P(1,1,1...,1,n+1)=0. This proves our inequality.

$\boxed{\text{Notation}}$ Define,$S_1=a_1+a_1a_2+\cdots +a_1\cdots a_{n-1}+2$. $S_2=a_2+a_2a_3+\cdots +a_2\cdots a_{n-1}+2$, In general ,$S_i =a_i +\cdots + a_i\cdots a_{n-1}+2$. $\forall i< n$. $\boxed{\text{Claim(1)}}$ $S_1 <2^{n-1}$. Proof Using Tchevbycev inequality we have , $\frac{S_1}{n} \le \frac {(n-1)a_1+2}{n} \frac{S_2}{n}$. $\implies S_1 \le \frac {(n-1)a_1+2}{n} S_2$. $\le \frac {(n-1)a_1+2}{n} . \frac{ (n-2)a_2+2}{n-1} S_3$. $\le \prod _{i=1}^{n+1} \frac {(n-i)a_i +2}{(n+1-i)}$.[Applying Induction on successive $S_i$]. $=\prod_{i=1}^{n-1} (a_i +\frac{2-a_i}{n+1-i})$. $< \prod_{i=1}^{n-1} (a_i +\frac {2-a_i}{2}$. $=\prod_{i=1}^{n-1} \frac {a_i+2}{2}$. $\le ( \frac{\frac{a_1+\cdots +a_{n-1}}{2}+(n-1)}{n-1})^{n-1}$.[using GM-AM] $<2^{n-1}$.

$\boxed{\text{Claim(2)}}$ $2^{n-1}< a_1\cdots a_n \le 2^{n}$. Proof The RHS inequality is trivial by AM-GM inequality. For LHS inequality I would like to use induction. $\boxed{n=2}$. We have $a_1+a_2=4$ ,$1<a_1\le 2$ and $a_2\ge 2$. $\implies a_1a_2 > 2$. Suppose , the statement is true for$n=k$ such that , $a_1+\cdots +a_k =2k$ and $a_1 \cdots a_K>2^{k-1}$. Now , consider $1<b_1 \le b_2 \cdots \le b_{k+1}$. Suppose ,$b_j =2$ is median of the sequence, and$a_i=b_i ,\forall i<j$ and $a_i=b_{i+1},\forall k\ge i>j$. $\implies b_1+...+b_{k+1}=2k+2$ and $b_1..b_{k+1}>2^k$. Our induction step is complete. This two claim leads$S_1<a_i \cdots a_n$ and equality for $a_1=a_2=\cdots =a_n=2$.

BOBTHEGR8 wrote: I think base case is $n=2$ which requires more or less same idea as the induction step and hence at the most $3$ marks will be deducted. But if your induction proof is totally different I don't know. Nope! Base case is $n=3$ In the problem it is written to prove for all $n\geq 3$...Maybe a typo in #1

Yes base case is 3 ....I also agree

For $n=2$, we want to show that $$a_1 + 2 \le a_1a_2$$where $a_1+a_2 = 4$ and $1 < a_1 \le a_2$. This is equivalent to showing that $(a_1-1)(a_1-2) \le 0$, which is true. Suppose, now, that the given inequality is true for $n=k$, where $k \ge 2$. Now, consider $k+1$ reals $1 < a_1 \le a_2 \le \cdots \le a_{k+1}$ with sum $2k+2$. Then, $1 < a_1 \le a_2 \le \cdots \le a_{k-1} \le a_k + a_{k+1}-2$ and $a_1 + a_2 + \cdots + a_{k-1}+(a_k+a_{k+1}-2) = 2k$, so by induction hypothesis, $$\sum_{i=1}^{k-1} a_1a_2 \cdots a_i + 2 \le a_1a_2 \cdots a_{k-1} \cdot (a_k + a_{k+1} -2)$$This means $$\sum_{i=1}^k a_1a_2 \cdots a_i + 2 \le a_1a_2 \cdots a_{k-1} \cdot (2a_k + a_{k+1} - 2) = a_1a_2 \cdots a_{k-1} \cdot (a_ka_{k+1}-(a_k-1)(a_{k+1}-2)) \le a_1a_2 \cdots a_{k-1} \cdot (a_ka_{k+1})$$or $$a_1a_2 \cdots a_k + a_1a_2 \cdots a_{k-1} + \cdots + a_1a_2 + a_1 + 2 \leq a_1a_2 \cdots a_{k+1}$$as desired. Personally, I found this the most difficult problem in the paper

My favorite in this year INMO. For convenience, let $P_i = a_1a_2\hdots a_i$. Apply Chevbyshev's inequality on $1<P_1<P_2<\hdots<P_{n-1}$ and $a_1\leq a_2\leq\hdots\leq a_n$ gives \begin{align*} (P_1+P_2+\hdots+P_n)&\geq\frac{1}{n}(a_1+a_2+\hdots+a_n)(1+P_1+P_2+\hdots+P_{n-1}) \\ &= 2(1+P_1+P_2+\hdots+P_{n-1}) \end{align*}which a simple rearranging makes this equivalent to the conclusion. Motivational Remark: The use of Chevbyshev's inequality is quite magical. Here are some (random) motivations I have before realizing it. The equality case $(2,2,\hdots,2)$. Moreover, if we allow $a_1=1$ (which I don't know why they put $1<a_1$), we get another equality case $(1,1,\hdots,1,n+1)$. The inequality is almost "linear". There are lots of "rearrangement" conditions.

Wow. I could never have thought of this, because it is counter-intuitive ( $P_1+P_2+\cdots + P_{n-1}$ is on the smaller side in the inequality).

anantmudgal09 wrote: Let $n \geqslant 2$ be an integer and let $1<a_1 \le a_2 \le \dots \le a_n$ be $n$ real numbers such that $a_1+a_2+\dots+a_n=2n$. Prove that$$a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n-2}+\dots+a_1a_2+a_1+2 \leqslant a_1a_2\dots a_n.$$ Proposed by Kapil Pause A small typo I would like to point out. It's actually $n\geq 3$ in the paper . Thanks And @2above, as @above pointed out there is a small typo in your solution

pranbunny wrote: What if we write 1/ai≤1/a1 Then apply the concept of GP.. and then get inequality in a1 only and check the minimum value at a1=a2=...=an=2? Anyone?

$\boxed {n=2} $ we have $a_1+2\leqslant a_1(\bullet)$ for $1\le a_1\le a_2\implies a_1\le 2$ and $a_1+a_2=4$ From $(\bullet) $ $\implies a_1+2\le a_1 (4-a_1)\implies(a_1-1)(a_2-2)\le 0$ is true Consider, $a_1a_2\cdots a_{n-2}+\cdots+a_1+2\leqslant a_1a_2\cdots <a_{n-1} $ for $1\leqslant a_1\leqslant\cdots\leqslant a_{n-1} $ and $a_1+\cdots+a_{n-1}=2 (n-1) $ Equality holds for $a_1=\cdots=a_{n-2}=1$ or, $a_1=\cdots=a_{n-1}=2$ where $a_{n-1}=n$ 1st Case Consider $a_1, a_2,\cdots,a_{n-1},a_n $ for which $a_1+a_2+\cdots+a_{n-1}+a_n=2n $ in this case $a_n\geqslant 2$ 2nd Case $a_{n-1}\left (a_{n-1}\right)\ge a_{n-1} $ Also, $$a_1+a_2+\cdots+a_{n-2}+a_{n-1}\left(a_{n-1}\right) =2n-\left (a_{n-1}+a_n\right)+a_{n-1}a_n-a_{n-1} =2n-2+\left (a_{n-2}\right)\left (a_{n-1}\right)\ge 2n-2$$Applying induction hypothesis for $a_1,\cdots,a_{n-2} $, 2nd case $a_{n-1}\left (a_{n-1}\right)=n-1$ Note, if we prove $a_1+\cdots+a_{n-1}=2 (n-1) $ then it is true for $a_1+\cdots +a_{n-1}\geqslant 2 (n-1)$ $$a_1+\cdots+a_{n-2}+a_1\cdots a_{n-3}+\cdots+a_1+2\leqslant a_1\cdots a_{n-2}\left (a_{n-1} a_n-a_{n-1}\right)\implies a_1\cdots a_{n-1}+a_1\cdots a_{n-2}+\cdots+a_1+2\leqslant a_1a_2\cdots a_{n-2}a_{n-1}a_n $$Equality holds for $$a_1=\cdots=a_{n-2}=1,a_{n-1}\left (a_{n-1}\right)=n-1 \implies a_1=\cdots=a_{n-2}=1,a_{n-1}\left (a_{n-1}\right)=2 \implies a_1=\cdots=a_{n-2}=a_{n-1}-1,a_n=n \implies a_1=\cdots=a_{n-1}=a_n=2$$

Supercali wrote: I think the inequality can be easily solved using induction on $n$ and applying the induction hypothesis to $a_1,a_2, \dots, a_{n-2}, a_n +a_{n-1} -2$. Yes, but we should have been careful while applying this. For those who solved it by induction. It should be done like this We prove the problem for stronger constraint $\sum a_i \ge 2n$. Base case $n=2$ is clear as $(a_2-2)(a_1-1)\ge 0\implies a_2a_1\ge 2a_1+a_2-2\ge 2+a_1$ as $a_1+a_2\ge 4$. Now assume it holds till $n-1$. We have $a_{n-1}(a_n-1)\ge a_{n-1} \ge a_{n-2} \cdots \ge a_1$. Also $a_1+ \cdots a_{n-2} + a_{n-1}(a_n-1)\ge 2n-a_{n-1}-a_n+a_{n-1}(a_n-1)=2n-2+(a_n-2)(a_{n-1}-1)\ge 2n-2$ We can apply induction hypothesis to $a_1, \cdots a_{n-2}, a_{n-1}(a_n-1)$ we get $a_1\cdots a_{n-2}+\cdots a_1+2\le a_1\cdots a_{n-2}a_{n-1}(a_n-1)$. Thanks to the proposer for pointing this out.

Hey, I got an easier solution to this, can anyone say if there is something wrong, For $n=2$, It reduces to $(a_1-1)(a_1-2) \leq 0$ which is true. Let's assume that it is true for $n=k$, then we will prove that it is true for $n=k+1$, It is true for $n = k \implies a_1a_2...a_{k-1}+ a_1a_2...a_{k-2}+......+a_1a_2 + a_1 + 2 \leq a_1a_2...a_{k}$ We need to prove that $a_1a_2...a_{k} + a_1a_2...a_{k-1}+......+a_1a_2 + a_1 + 2 \leq a_1a_2...a_{k+1}$, We substitute that $a_1a_2...a_{k-1}+ a_1a_2...a_{k-2}+......+a_1a_2 + a_1 + 2 \leq a_1a_2...a_k$ to get, $ a_1a_2...a_{k}+ a_1a_2...a_{k-1}+......+a_1a_2 + a_1 + 2 \leq 2a_1a_2...a_n \leq a_1a_2...a_ka_{k+1}$, as $a_{k+1} \geq 2$, Hence Proved . Wrong @below

Krish230905 wrote: We substitute that $a_1a_2...a_{k-1}+ a_1a_2...a_{k-2}+......+a_1a_2 + a_1 + 2 \leq a_1a_2...a_k$ to get. It's wrong. You are inducting with constraint $a_1+a_2...+a_{k+1}=2k+2$ and above statement holds when $a_1+a_2...+a_{k}=2k$

kapilpavase wrote: Krish230905 wrote: We substitute that $a_1a_2...a_{k-1}+ a_1a_2...a_{k-2}+......+a_1a_2 + a_1 + 2 \leq a_1a_2...a_k$ to get. It's wrong. You are inducting with constraint $a_1+a_2...+a_{k+1}=2k+2$ and above statement holds when $a_1+a_2...+a_{k}=2k$ But isn't it supposed to be $2k+2$ for $n=k+1$

For different $k$, not only the inequalities are different, but constraints are also different. So you cannot substitute result with $n=k$ to prove result for $n=k+1$, like you did above.

My Solution: we can use induction here so lets see the base case here at $n=2$ we have, \[a_1+a_2=4,a_1\geq a_2\text{ and }a_1,a_2\geq 1 \]we have to prove the base case which is \[2+a_1\leq a_1a_2\implies 2+a_1\leq a_1(4-a_1)\implies 2+a_1\leq 4a_1-a_1^2\]\[a_1^2-3a_1+2\leq 0\implies a_1^2-2a_1-a_1+2\leq 0\implies a_1(a_1-2)-1(a_1-2)\leq 0\]\[(a_1-1)(a_1-2)\leq 0\]which is basically true so lets assume that this true for some $n$.we have prove this for some $n+1$ \[a_1+a_2+\dots+a_n+a_{n+1}=2(n+1),a_1\leq a_2\leq\dots\leq a_n\leq a_{n+1}\text{ and } \]\[a_1,a_2,\dots,a_na_{n+1}\geq 1\]so we have to prove that \[\implies 2+a_1+a_1a_2+\dots+a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n}\leq a_1a_2\dots a_na_{n+1}\]we have \[a_1+a_2+\dots+a_n+a_{n+1}=2n+2\implies a+1+a_2+\dots+\underbrace{a_n+a_{n+1}-2}_{a_n'}=2n\]so we can see here that $a_{n+1}\geq 2$ since $a_{n+1}\geq a_n\geq\dots\geq a_2\geq a_1$ so we can say that \[a_n'=a_n+a_{n+1}-2\geq a_n\geq a_{n-1}\geq \dots\geq a_2\geq a_1\]now we can apply induction on $a_1,a_2,a_3,\dots,a_{n-1},a_n'$ so according to the condition we have \[2+a_1+a_1a_2+\dots+a_1a_2\dots a_{n-2}+a_1a_2\dots a_{n-1}\leq a_1a_2\dots a_n'\]by adding $a_1a_2\dots a_n$ on both side we have \[2+a_1+a_1a_2+\dots+a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n}\leq a_1a_2\dots a_{n-1}a_n'+a_1a_2\dots a_{n}\]so its suffice to prove that \[a_1a_2a_3\dots a_{n-1}(a_n+a_{n+1}-2)+a_1a_2a_3\dots a_n\leq a_1a_2a_2\dots a_{n+1}\]taking $a_1a_2\dots a_{n-1}$ common on both side and we have \[a_1a_2\dots a_{n-1}(a_n+a_{n+1}-2+a_n)\leq a_1a_2\dots a_{n+1}\]\[\implies 2a_n+a_{n+1}-2\leq a_na_{n+1}\implies a_na_{n+1}-a_{n+1}-2a_n+2\geq 0\]\[\implies a_{n+1}(a_n-1)-2(a_n-1)\geq 0\implies (a_n-1)(a_{n+1}-2)\geq 0\]so we can see here that $a_n> 1$ and $a_{n+1}\geq 2$ which is basically true since $a_{n+1}$ is the largest and we are done!