# kshitij's latest activity

• I didn't knew there were multiple winning strategies for ##P##! Also, couldn't we generalise that further for when ##P## chooses any...
• kshitij reacted to fresh_42's post in the thread Challenge Math Challenge - July 2021 with Like.
Ok, I finally got it. It took so long and several posts that I post my solution as reference: ##P## has the following winning strategy...
• When I first used ##A.M \geq G.M## on$$f(x,y,z)=c(c-2x)(c-2y)(c-2z)$$ the equality would never hold because if $$c=c-2x=c-2y=c-2z$$ then...
• So, basically as I understand it, we had $$f(x,y,z)=c(c-2x)(c-2y)(c-2z)$$ And ##c## is a constant, so we use ##A.M \geq G.M## on...
• kshitij reacted to fresh_42's post in the thread Challenge Math Challenge - July 2021 with Like.
I am saying that any upper bound is not sufficient. We are looking for the maximum. We get from your calculation and the fact that...
• Also I forgot that ##x,y,z \gt 0## so ##x=y=z=0## cannot be correct anyway. I will try again this problem later.
• But if ##f(x,y,z)## is area of a triangle then ##f(x,y,z)=0## is the least possible value not the maximum value, surely this is wrong!
• Are you saying that I must prove why the equality holds? or do you want me to show when the equality holds? As I said that for...
• I think I have just a strategy but I am not sure because lets say that P plays according to my strategy and chooses a negative value of...
• I just noticed that $$c(c-2x)(c-2y)(c-2z)$$ Looks similar to the formula for area of a triangle! Is that what you meant, is geometrical...
• I don't know about why is that maximum, if equality holds then they should be maximum. I think I don't understand what you're trying to...
• I don't understand what you're saying? How can the equation ##y=x^3+c## have three different real roots for any value of ##c##?
• I might be wrong, but I thought that when ##a=0## then, if the value of ##b## is chosen to be positive, then the equation...
• But I applied ##A.M \geq G.M## on $$(z-x+y)(z+x-y)(x+y-z)(x+y+z)$$ ##(z-x+y),(z+x-y),(x+y-z)## and ##(x+y+z)## aren't necessarily...
• You can ignore this attempt, as the question doesn't say that ##x,y,z\gt 0## so we cannot use ##A.M \geq G.M##