Maximum of $f(x,y)=\frac{1}{x+y+1}-\frac{1}{(x+1)(y+1)}$ attained by $(x,y)=(\frac{1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})$ and maximum value is $$\frac{11+5\sqrt{5}}{2}$$Solution : After some Algebraic Manipulation we need to prove that $$\frac{11+5\sqrt{5}}{2}xy \leqslant (x+y+1)(x+1)(y+1)$$Let $xy=k^2$ so there are exist $a,b \in \mathbb{R}^{+}$ such that $$x=\frac{ka}{b} , y=\frac{kb}{a}$$and inequality becomes , $$\frac{11+5\sqrt{5}}{2}abk^2 \leqslant (ka^2+kb^2+ab)(ka+b)(kb+a)$$But from Cauchy-Schwarz inequality and AM-GM We deduce that $$(ka+b)(kb+a) \geqslant (k+1)^2ab , ka^2+kb^2 \geqslant 2ab$$So we need to prove that $$(2k+1)(k+1)^2-\frac{11+5\sqrt{5}}{2}k^2 \geqslant 0$$for all $k \in \mathbb{R}^{+}$ which is easy.

I think it's $\dfrac{-11+5\sqrt{5}}{2}$, you get a typo. $LHS \leq \dfrac{1}{x+y+1}-\dfrac{4}{(x+y+2)^2}$ Let $x+y+2=t>2 \Rightarrow LHS \leq f(t)=\dfrac{1}{t-1}-\dfrac{4}{t^2}$ for $t>2$ $f'(t)=\dfrac{(2-t)(t^2-6t+4)}{(t-1)^2t^3}=0 \Rightarrow t=3+\sqrt{5}$ Easy to complete that $f(t)_{max}=f(3+\sqrt{5})=\dfrac{-11+5\sqrt{5}}{2}<\dfrac{1}{11}$

parmenides51 wrote: Prove that $\frac{1}{x+y+1}-\frac{1}{(x+1)(y+1)}<\frac{1}{11}$ for all positive $x$ and $y$. Uzbekistan 2015

parmenides51 wrote: Prove that $$\frac{1}{x+y+1}-\frac{1}{(x+1)(y+1)}<\frac{1}{11}$$for all positive $x$ and $y$. Prove that for all positive real numbers $x, y$ and $z$, the double inequality $$0 < \frac{1}{x + y + z + 1} -\frac{1}{(x + 1)(y + 1)(z + 1)} \le \frac18$$holds. Kameawtamprooz wrote: we need to prove that $$\frac{11+5\sqrt{5}}{2}xy \leqslant (x+y+1)(x+1)(y+1)$$ Let $a,b,c$ be positive real numbers . Prove that$$(a + b)(b + c)(a+b+c)\geq\frac {5\sqrt 5+11} {2}abc.$$

Let $x,y,z$ be positive real numbers . Prove that$$\frac{1}{x+y+z^2}-\frac{1}{(x+z)(y+z)}\leq\frac {5\sqrt 5-11} {2}$$z=1

sqing wrote: Let $x,y,z$ be positive real numbers . Prove that$$\frac{1}{x+y+z^2}-\frac{1}{(x+z)(y+z)}\leq\frac {5\sqrt 5-11} {2}$$

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