<s>[INST] Let $(x,y) \in \Bbb R^+$ Prove that $\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$ Let $(x,y) \in \Bbb R^+$ Prove that $$\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$$
My try
Well, i didn't see a way to factorize this, so i put it in WolframAlpha and i got that we can rewrite it like this, which is obviously true for $(x,y) \in \Bbb R^+$:
$$\frac{(y-x)^2(x+y+1)}{xy(x+y)^2}\ge 0$$
But i don't see the way to get there manually, im stuck here:
$$\frac{(x+y)(y+1)}{xy}\ge \frac{(x+y+2)^2}{(x+y)^2}$$
 [/INST] it is just $AM-GM$
multiplying out we get
$$(x+y)^2+\frac{(x+y)^2}{x}+\frac{(x+y)^2}{y}+\frac{(x+y)^2}{xy}\geq (x+y)^2+4+4(x+y)$$ and this is
$$xy(x+y)^2+y(x+y)^2+x(x+y)^2+(x+y)^2\geq xy(x+y)^2+4xy+4xy(x+y)$$
$$(x+y)^3+(x+y)^2\geq 4xy+4xy(x+y)$$
and this is $$(x+y)^2(x+y+1)\geq 4xy(x+y+1)$$
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