fattypiggy123 wrote: Given that $0 < x,y < 1$, find the maximum value of $\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)}$ From the result for all $a,b,c$ in $\Bbb{R+}$ $\frac {abc}{(a+b)(b+c)(c+a)}\leq \frac{1}{8} $ we get solution easily by setting $a=x,b=y,c=1-x-y$ So maximum value is $\frac{1}{8}$ and it is reached iff $a=b=\frac {1}{3} $.

Let $S=x+y$, $P=xy$. Consider: $f(P)$. We have $f'(P)>0$, hence $f(P)\leq{f(\frac{S^2}{4})}$, continue..

Let $z=1-x-y .$ When $x+y\geq 1 ,$ $\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)}\leq 0<\frac{1}{8};$ When $x+y<1 ,$ $\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)}=\frac{xyz}{(x+y)(y+z)(z+x)}\leq\frac{xyz}{2\sqrt{xy}\cdot2\sqrt{yz}\cdot2\sqrt{zx}}=\frac{1}{8}.$ Hence , $\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)}\leq \frac{1}{8}.$ Equality holds for $ x=y=\frac{1}{3}.$

This is a specific case of 1998 ISL A1 https://artofproblemsolving.com/community/c6h18486p124417

Obviously if $1-x-y < 0$, the expression is negative. Else, substitute $z = 1-x-y$, so $$\frac{xyz}{(x+y)(y+z)(z+x)} \leq \frac 18$$by AM-GM.

Let $x,y \in (0,1) .$ Prove that $$\frac{xy(2-3x-3y)}{(x+y)(1-x)(1-y)}\leq \frac{1}{8}$$Equality holds when $x=y=\frac{1}{5}.$

HamstPan38825 wrote: Obviously if $1-x-y < 0$, the expression is negative. Else, substitute $z = 1-x-y$, so $$\frac{xyz}{(x+y)(y+z)(z+x)} \leq \frac 18$$by AM-GM. Bravo

HamstPan38825 wrote: Obviously if $1-x-y < 0$, the expression is negative. Else, substitute $z = 1-x-y$, so $$\frac{xyz}{(x+y)(y+z)(z+x)} \leq \frac 18$$by AM-GM. See 5#

AnhIsGod wrote: Let $S=x+y$, $P=xy$. Consider: $f(P)$. We have $f'(P)>0$, hence $f(P)\leq{f(\frac{S^2}{4})}$, continue.. Indeed

Let $x,y \in (0,1) .$ Prove that $$\frac{xy(1-x-y)}{(x+y)(k-x)(k-y)}\leq \frac{1}{8k(2k-1)}$$Where $k\geq 1.$

S

$$\frac{xy(1-x-y)}{(x+y)(2-x)(2-y)}\leq \frac{1}{48}$$

sqing wrote: Let $x,y \in (0,1) .$ Prove that $$\frac{xy(1-x-y)}{(x+y)(k-x)(k-y)}\leq \frac{1}{8k(2k-1)}$$Where $k\geq 1.$

S

$$\frac{xy(1-x-y)}{(x+y)(2-x)(2-y)}\leq \frac{1}{48}$$ Are you sure?

mihaig wrote: sqing wrote: Let $x,y \in (0,1) .$ Prove that $$\frac{xy(1-x-y)}{(x+y)(k-x)(k-y)}\leq \frac{1}{8k(2k-1)}$$Where $k\geq 1.$

S

$$\frac{xy(1-x-y)}{(x+y)(2-x)(2-y)}\leq \frac{1}{48}$$ Are you sure? fattypiggy123 wrote: Given that $0 < x,y < 1$, find the maximum value of $\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)}$

Attachments:

sqing wrote: mihaig wrote: sqing wrote: Let $x,y \in (0,1) .$ Prove that $$\frac{xy(1-x-y)}{(x+y)(k-x)(k-y)}\leq \frac{1}{8k(2k-1)}$$Where $k\geq 1.$

S

$$\frac{xy(1-x-y)}{(x+y)(2-x)(2-y)}\leq \frac{1}{48}$$ Are you sure? The quoted message was pointless