<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] I give a proof using Cauchy-Schwartz for fun.
$$\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$$
$$\iff \Bigr((x+y)(1+\frac{1}{x})\Bigl)\Bigr((x+y)(1+\frac{1}{y})\Bigl)\ge \Bigr(x+y+2\Bigl)^2$$
$$\iff \Bigr(x+y+1+\frac{y}{x}\Bigl) \Bigr(x+y+1+\frac{x}{y}\Bigl) \ge \Bigr(x+y+2\Bigl)^2$$
$$\iff \Bigr(\sqrt x^2+\sqrt y^2+1^2+\sqrt{\frac{y}{x}}^2\Bigl) \Bigr(\sqrt x^2+\sqrt y^2+1^2+\sqrt{\frac{x}{y}}^2\Bigl) \\
\ge (\sqrt x \cdot \sqrt x + \sqrt y \cdot \sqrt y + 1\cdot1 + \sqrt{\frac{y}{x}} \cdot \sqrt{\frac{x}{y}})^2 $$
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