shobber wrote: Let $a_1,a_2,a_3,\ldots,a_n$ be a sequence of non-negative integers, where $n$ is a positive integer. Let \[ A_n={a_1+a_2+\cdots+a_n\over n}\ . \] Prove that \[ a_1!a_2!\ldots a_n!\ge\left(\lfloor A_n\rfloor !\right)^n \] where $\lfloor A_n\rfloor$ is the greatest integer less than or equal to $A_n$, and $a!=1\times 2\times\cdots\times a$ for $a\ge 1$(and $0!=1$). When does equality hold? http://www.kalva.demon.co.uk/apmo/asoln/asol021.html

I don't know why I can't open all the links from this website.

try use proxy.

Kalva is not working anymore. Can someone post a solution? I'm stucked. I have tried induction but failed.

There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) = \log x!$ It is understandable because $ \log x! = \log 1 + \log 2 + \cdots \log x$ and $ \log x$ is an increasing function. However I dont see a rigorous proof now. Also it can be made monotonically increasing(not strictly) by making $ f(x) = 0\ \forall x\geq 0$ and $ \leq 1$ $ \sum_{i = 1}^{n}f(a_n)\geq nf(A_n)$ by Jensen's inequality $ f(A_n)\geq f(\lfloor A_n \rfloor)$ because $ f(x)$ is monotonically increasing. $ \sum_{i = 1}^{n}f(a_n) = \sum_{i = 1}^{n}\log a_n! = \log\prod a_n!$ $ nf(\lfloor A_n \rfloor) = n\log \lfloor A_n \rfloor ! = \log (\lfloor A_n \rfloor !)^n$ Thus proved. Equality holds for $ a_1 = a_2 = \cdots = a_n$ or when some of $ a_i$ are $ 1$ and some are $ 0$.

Kalva still lives in our hearts... and someplace else (Webarchive). So here's a link to the solution: http://web.archive.org/web/20040313130703/http://www.kalva.demon.co.uk/apmo/asoln/asol021.html

Great solution in Kalva! But I can't help saying that the equality case given there is not truly an "iff" statement

I think those archive is a little bit different from the real kalva. For example, I didn't see tournament of the towns in kalva before (am I blind?)

Akashnil wrote: There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) = \log x!$ $ f(x)=\log x!$ is concave!

Sunjee wrote: Akashnil wrote: There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) = \log x!$ $ f(x) = \log x!$ is concave! Honestly, terms convex and concave always seem to be so confusing: in different literature, I used to find diametrically opposite definitions: in some books I've used, what is concerned concave is convex in others: so some use the terms concave up and concave down- now that's a little more unambiguous.

Sunjee wrote: Akashnil wrote: There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) = \log x!$ $ f(x) = \log x!$ is concave! For some function defined on reals, I mean it to be convex iff the 2nd derivative is always non-negetive. And I'm pretty sure there is a convex function $ f(x)$ which has the required property. Edit: Can we call a sequence convex? In that case the definition would be $ a_{n - 1} + a_{n + 1}\geq 2a_n$. Jensen possibly holds there.

Akashnil wrote: There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) = \log x!$ It is understandable because $ \log x! = \log 1 + \log 2 + \cdots \log x$ and $ \log x$ is an increasing function. However I dont see a rigorous proof now. Correct! In fact, there is exactly one such function; it is the logarithm of the gamma function (shifted by one). Now: I actually think that the right-hand-side divides the left-hand-side under suitable conditions, but I'm not sure what they are.

I don't understand in kalva web ! Could anyone explain it? Thanks.

Just start with $ a_1=a_2...=a$ Then clearly we have equality. Now think about what happens when we increase one of the $ a_i$ by one, and decrease another by $ 1$. Then we are dividing the $ LHS$ by $ a$, and multiplying by $ a+1$. In this way we see any change preserves the inequality.

Akashnil wrote: There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) = \log x!$ It is understandable because $ \log x! = \log 1 + \log 2 + \cdots \log x$ and $ \log x$ is an increasing function. However I dont see a rigorous proof now. . Since $ \log (x+1)! - \log x! = log (x+1) \geq 0$ so $ f(x)= \log x!$ is increasing!

I came up with a solution. Does it work? Solution Let $ \lfloor A_n \rfloor = p$. Let $ b_m = a_m - p$ for all $ a_m \ge p$ and let $ c_k = p - a_k$ for all $ a_k \le p$. Now note that since $ \sum a_i \ge np$, it follows that $ \sum b_m \ge \sum c_k$. Now consider the following product. Define that, $ P = \frac{(p+1)(p+2) \dots (p+b_1)(p+1)(p+2) \dots (a+b_m)}{(p-c_1+1)(p-c_1+2) \dots (p)(p-c_2+1)(p-c_2+2) \dots (p-c_k+1) \dots (p)}$ There are a total of $ \sum b_m$ factors in the numerator each of which is greater than or equal to $ p$ and there are a total of $ \sum c_k$ factors in the denominator each of which is less than or equal to $ p$ (since it is possible that for some $ i$, $ b_i$ or $ c_i$ is $ 0$). Since $ \sum b_m \ge \sum c_k$, it follows that $ P \ge 1$. Now note that $ P = \frac{a_1!}{p!} \cdot \frac{a_2!}{p!} \cdot \dots \cdot \frac{a_n!}{p!}$ Hence since $ P \ge 1$, it follows that $ a_1!a_2! \dots a_n! \ge (p!)^n$. Equality holds when $ P = 1$ which can only occur if all $ c_i$ and $ b_i$ are $ 0$ which occurs when $ a_1 = a_2 = \dots = p$.

For all $a_i$ being equal the equality case is evident. Now assume that all $a_i$ are not equal, and without loss of generality let $a_1 \leqslant a_2 \leqslant ... \leqslant a_n$. Then for some $x$ such that $a_n \geqslant a_1 +x$, the transformation $(a_1, a_2,...,a_n)\mapsto (a_1 +x, a_2, ...,a_n-x)$ decreases the left hand side. Clearly, the right hand side remains invariant under this transformation. The value of the left hand side decreases by every such transformation. This transformation cannot be performed when all the numbers are equal, hence the minimum value of the left hand side is when all $a_i$ are equal. That, as explained above, gives the equality case. Hence we're done.

Notice that by taking $\ln$ of both sides, the inequality becomes: $\sum_{i=1}^{n}\ln {a_i!}\ge\ln {(\lfloor A_n\rfloor)^n}$ Furthermore let $f(x)=\ln {x!}\Longrightarrow f'(x)=\psi(x+1)\Longrightarrow f''(x)=\psi_1(x+1)>0$ thus the function is strictly convex on the non-negative integers. Thus $\sum_{i=1}^{n}\ln{a_i!}=\sum_{i=1}^{n}f(a_i)\overset{\text{Jensen}}{\ge}n f\bigg(\frac{\sum_{i=1}^{n}a_i}{n}\bigg)=nf(A_n)=n\ln {A_n!}$ Furthermore: $n\ln {A_n!}\ge\ln {(\lfloor A_n\rfloor)^n}\Longleftrightarrow (e^{\ln {A_n!}})^n\ge e^{\ln {(\lfloor A_n\rfloor)^n}}\Longrightarrow (A_n)^n\ge(\lfloor A_n\rfloor)^n\Longrightarrow A_n!\ge\lfloor A_n\rfloor$ which is clearly true $\blacksquare$. And we are done!

let a_1>=a_2>=....a_n Now we will prove this by induction A_1=a_1 so p(1) is true now let p(n-1) be true and note p(n) is rewarding (a_n)!/([A_n-1]!)>=(([A_n])/([A_n-1]))^n and this is true because (a_n)-([A_n-1])=n(([A_n])-([A_n-1])) and a_n>=[A_n] and equality hold iff (a_i)!=(a_j)! for all 1<=(i,j)<=n