RockmanEX3 wrote: Let $x$, $y$ be positive real numbers with $xy = 4$. Prove that $$\frac{1}{x+3} + \frac{1}{y+3} \le \frac{2}{5}$$ For which $x$ and $y$ does equality hold? (Walther Janous) Substituting $y=4/x$ The expression simplifies to $(x-2)^{2}$ > or = 0 which is obviously true Equality holds at $(x,y)=(2,2)$

RockmanEX3 wrote: Let $x$, $y$ be positive real numbers with $xy = 4$. Prove that $$\frac{1}{x+3} + \frac{1}{y+3} \le \frac{2}{5}$$ For which $x$ and $y$ does equality hold? (Walther Janous) See here

See, $$\frac{1}{x+3} + \frac{1}{y+3} =\frac{x+y+6}{3(x+y)+13} \le \frac{2}{5}\iff x+y\geq4 =2\sqrt{xy}.$$With AM-GM, equality holds when $x=y=2$.

First, we substitute $x=\frac{4}{y}$ into the original expression; $$\frac{1}{\frac{4}{y}+3}+\frac{1}{y+3} \le \frac{2}{5}.$$ Expanding and simplifying, we have $$\frac{y+\frac{4}{y}+6}{4+3y+\frac{12}{y}+9} \le \frac{2}{5}$$ $$\frac{y^2+6y+4}{3y^2+13y+12} \le \frac{2}{5}$$ $$5y^2+30y+20 \le 6y^2+26y+24$$ $$y^2-4y+4 \ge 0$$ $$(y-2)^2 \ge 0.$$ This is true by the Trivial Inequality, and equality holds at $(x,y)=(2,2).$

Let $t=x+y$, then $\frac1{x+3}+\frac1{y+3}=\frac{t+6}{3t+13}$. We want $\frac{t+6}{3t+13}\le\frac25\Leftrightarrow t\ge4$, which is true since $t\ge2\sqrt{xy}$ by AM-GM.

Same as above oops

Let $a, b>0$ and $ ab= 4.$ Prove that $$\frac{1}{a+3}+\frac{1}{2b+3}\leq 6-4\sqrt{2}$$$$\frac{1}{a+3}+\frac{2}{b+3}\leq \frac{9-4\sqrt{2}}{5}$$

sqing wrote: Let $a, b>0$ and $ ab= 4.$ Prove that $$\frac{1}{a+3}+\frac{1}{2b+3}\leq 6-4\sqrt{2}$$