A Challenging Inequality, 2024 UMD Competition, Part II, Problem 4

[@sibidinakaran4757]

Dr. Ebrahimian loved having you for 340 last year!

[@DrEbrahimian]

That was a great class!

[@surajmandal_567]

Hi doctor I am having problem proving the following identity: For 1

[@DrEbrahimian]

I suggest looking at small cases on the difference of m and r. See if you can prove the equality for m-r=0, 1, 2, etc. I suggest you try proving it with little algebra. I haven’t solved it though.

[@huyviethungnguyen7788]

I think using strong induction for m+r and the fact that xCy=(x-1)C(y-1)+(x-1)Cy would work.

[@DrEbrahimian]

This is an interesting idea. Thank you for sharing. I don’t know that this idea would work. You end up getting a different identity that needs to be proved.

[@huyviethungnguyen7788]

@@DrEbrahimianLet's denote the sum as f(r,m). I split (k-1)C(r-1) × mCk into (k-2)C(r-2) × (m-1)C(k-1) + (k-2)C(r-1) × (m-1)C(k-1) + (k-1)C(r-1) × (m-1)Ck), which enables me to write f(r,m) in terms of f(r-1,m-1), f(r,m-1). That doesn't work? I haven't done the work to check tho because I don't have paper with me right now.

[@DrEbrahimian]

@ I don’t know that the second sum can be written in terms of f(r, m-1).