Final Round

By the Cauchy inequality, $$(x+1)(y+1)\le \frac{((x+1)+(y+1){)}^2}{4}=\frac{(x+y+2{)}^2}{4}.$$ So the required inequality $$\frac{1}{x+y+1}-\frac{1}{(x+1)(y+1)}<\frac{1}{11},\text{if }x>0\text{ and }y>0,(1)$$ is a consequence of the inequality $$\frac{1}{x+y+1}-\frac{4}{(x+y+2{)}^2}<\frac{1}{11},\text{if }x>0\text{ and }y>0.(2)$$ Putting $x+y+1=t$ ($t>1$, since $x$ and $y$ are positive) into (2), we obtain the inequality $$\frac{1}{t}-\frac{4}{(t+1{)}^2}<\frac{1}{11},\text{if }t>1(3)$$ which is equivalent to (2). This inequality can be represented as $$t^3-9t^2+23t-11>0,\text{if }t>1.(4)$$ Inequality (1) will be proved once we prove (4). We rearrange (4) as $$t^3-9t^2+23t-11=(t-3{)}^3-4t+16=(t-3{)}^3-4(t-3)+4=(t-1)(t-3)(t-5)+4.$$ Therefore, (4) is equivalent to the inequality $$(t-1)(t-3)(t-5)+4>0,\text{if }t>1.(5)$$ We see that for $t>1$ the polynomial $(t-1)(t-3)(t-5)\le 0$ only if $t\in [3;5]$. But in this case $0<t-1\le 4$ and $(t-3)(t-5)\ge -1$, therefore, $(t-1)(t-3)(t-5)\ge -4$ for $t\in [3;5]$. Thus (5) is proved, which implies that (1) holds.