Any ideas?

Hi where do you See this question I think it is not Azerbaijan I am from Azerbaijan.

It is JBMO SL problem

den_thewhitelion wrote: Any ideas? You can see here: https://artofproblemsolving.com/community/u295258h1239668p6322741

den_thewhitelion wrote: Let $x ,y, z$ be real numbers, satisfying the relations $x \ge 20$ ,$y \ge 40$ ,$z \ge 1675$ and $x + y + z = 2015$ .Find the greatest value of the product $P = xy z$ JBMO Shortlist 2015 , A1(Moldova) $ (1675-z) (1675-y) \le 0 \implies yz \le 1675 (y + z-1675)$ . Hence $xyz \le 1675x (y + z-1675) \le 1675 (\frac {x + y + z-1675} {4}) ^ 2 = 1675 * 170 ^ 2$ . Equality holds when $z = 1675, x = y$

I have a different solution: xyz<=(x+y+z)^2/3^2= 2015^2/9= 451,136.111111 (AM-GM) So the maximum of xyz is the closest value to 451,136.111111 which means that max(xyz)=451,136

Okay I am pretty sure I messed this up but can someone tell me what happened.

Use Karamata at $ln(x) + ln (y) + ln (z)$

OlympusHero wrote: and we want to maximize $x_1y_1z_1$ What did you do wrong here?

Oops complete idiocy strikes again Will do it again.

Can I have a hint on how to continue?

Bumping this. Do I do something like $\frac{(x_1+20)+(y_1+40)+(z_1+1675)}{3} \geq \sqrt[3]{(x_1+20)(y_1+40)(z_1+1675)}$? But that's just the same as $\frac{x+y+z}{3} \geq \sqrt[3]{xyz}$ which is not at all helpful.

Yes, just simplify that expression.

Do you mean $(x_1+20)(y_1+40)(z_1+1675)$? Assuming you do, expanding gives something nasty that can only be simplified in one small way...

No, simplify the LHS.

Well we have $\frac{(x_1+20)+(y_1+40)+(z_1+1675)}{3}=\frac{2015}{3}$ so $\frac{2015}{3} \geq \sqrt[3]{(x_1+20)(y_1+40)(z_1+1675)}$ and we can just cube? Does that work?

OlympusHero wrote: Bumping this. Sorry, I know this seems pointless, but I feel like I'm doing something bogus and I'd like someone to check. check the equality cases

Yeah, the equality case is $x=y=z=\frac{2015}{3}$, but it doesn't satisfy the $z$ restriction. So how do I approach this now??

sttsmet wrote: I have a different solution: xyz<=(x+y+z)^2/3^2= 2015^2/9= 451,136.111111 (AM-GM) So the maximum of xyz is the closest value to 451,136.111111 which means that max(xyz)=451,136 Not achievable. OlympusHero wrote: Yeah, the equality case is $x=y=z=\frac{2015}{3}$, but it doesn't satisfy the $z$ restriction. So how do I approach this now?? So this method doesn't work.

jasperE3 wrote: So this method doesn't work. I see, thanks! Can you give a hint on what to do from here: OlympusHero wrote:

Can I have a hint on how to continue? Or maybe the original approach is wrong? I took it from here: HVishy wrote:

Bumping this.

Quote: Darkztar wrote: Use Karamata at $ln(x) + ln (y) + ln (z)$ Sorry for Bump, can someone show the Karamata solution

Allow me to correct my previous solution and make this problem clear for everyone Since $z \geq 1675$ set $z=1675+k$. Let y has a standard value that does not effect either the sum or the product. We will now replace $x$ with $x+k$ and $z$ with $1675$. Obviously the sum remains $x+y+1675+k=2015$ but $(x+k)1675 \geq x(1675+k)$ since $1675 \geq x$. Therefore $P$ is maximized when $z=1675$. Now in order to find $x, y$ we have that $x+y=340$ we apply $AM-GM$ from which we have that $xy \leq (\frac{x+y}{2})^2=(\frac{340}{2})^2=170^2$ which is achieved when $x=y=170$. So maximum $P$ is $1675*170^2$

Solution: wrote: $$x=20+a,\ y=40+b,\ z=1675+c,\ a,b,c\geq0,\ a+b+c=340.$$ $$xyz=(20+a)(40+b)(340-a-b)\\=(800+40a+20b)(340-a-b)\\\leq(800+40a+40b)(340-a-b)\\=40(20+a+b)(340-a-b)\\=40(6800+320(a+b)-(a+b)^2)\\=40(180^2 -(160-(a+b))^2)\leq40(180^2),$$Equality when $a+b=160,$ and $b=0,\ a=160.$ $\max\{xyz\}=40\times{180^2}$

Vulch wrote: Solution: wrote: $$x=20+a,\ y=40+b,\ z=1675+c,\ a,b,c\geq0,\ a+b+c=340.$$ $$xyz=(20+a)(40+b)(340-a-b)\\=(800+40a+20b)(340-a-b)\\\leq(800+40a+40b)(340-a-b)\\=40(20+a+b)(340-a-b)\\=40(6800+320(a+b)-(a+b)^2)\\=40(180^2 -(160-(a+b))^2)\leq40(180^2),$$Equality when $a+b=160,$ and $b=0,\ a=160.$ $\max\{xyz\}=40\times{180^2}$ Would anyone tell me where am I getting wrong?

can solve this but it's not considered a valid solution for olympiads.