Related problems should be 2017 Polish Final P6 and 2003 IMO SL A6.

Also related to this problem.But I don't think it's possible to apply the methods mentioned above to this problem.

Is this a discrete form of Prekopa-Leindler Inequality?

I roared for 4 hrs and finally made it to the weaker version of m=n... I'm so vegetable 我嚎了4个小时才整出来m=n的弱化情况…… 我太菜了

5th hr: okay if m-n is odd 第五个小时写出来m和n差为奇数时的情况

though I've managed it out,but it's quite lengthy

It sure is difficult

wrong $~~~$

Indeed, this problem has nothing to do with Prekopa-Leindler Inequality, and every solution I have seen so far which applies integral turns out to be a fake solve. Try to see this as a combinatorics problem.

Rushery_10 wrote: Indeed, this problem has nothing to do with Prekopa-Leindler Inequality, and every solution I have seen so far which applies integral turns out to be a fake solve. Try to see this as a combinatorics problem. Yep,I've been doing it in another way

The clever solution by contestant TJF: Induct on $m+n$: If $m=0$ or $n=0$, obvious. Suppose correct when $m+n-1$, and consider $m,n\ge1$. Let $A=\sum_{i=0}^{m}a_i,B=\sum_{j=0}^{n}b_j$. By IH, $\sum_{k=1}^{m+n}c_k\ge\frac{m+n}{m(n+1)}\sum_{i=1}^{m}a_i\sum_{j=0}^{n}b_j=\frac{m+n}{m(n+1)}(A-a_0)B$ $\sum_{k=1}^{m+n}c_k\ge\frac{m+n}{(m+1)n}\sum_{i=0}^{m}a_i\sum_{j=1}^{n}b_j=\frac{m+n}{(m+1)n}A(B-b_0)$. It suffices to prove that $a_0b_0+\max\left(\frac{m+n}{m(n+1)}(A-a_0)B,\frac{m+n}{(m+1)n}A(B-b_0)\right)\ge\frac{(m+n+1)AB}{(m+1)(n+1)}$. For simplicity, denote $x=\frac{a_0}A,y=\frac{b_0}B$. It suffices to prove that $(m+1)(n+1)xy+(m+n)\max\left(\frac{m+1}m(1-x),\frac{n+1}{n}(1-y)\right)\ge m+n+1$. Let $t=\max\left(\frac{m+1}m(1-x),\frac{n+1}{n}(1-y)\right)$. $x\ge1-\frac{mt}{m+1},y\ge1-\frac{nt}{n+1}$. It suffices to prove that $(m+1-mt)(n+1-nt)+(m+n)t\ge m+n+1$. But that's just $mn(t-1)^2\ge0$. $\Box$