<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's just Jensen for $f(x)=\ln\left(1+\frac{1}{x}\right)$.
Indeed, $f''(x)=\frac{(2x+1)}{x^2(x+1)^2}>0$ and we get
$$\frac{\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\frac{1}{x}\right)}{2}\geq\ln\left(1+\frac{1}{\frac{x+y}{2}}\right),$$
which is your inequality.
Also, it's
$$1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\geq1+\frac{4}{x+y}+\frac{4}{(x+y)^2},$$ which is true by C-S
$$\frac{1}{x}+\frac{1}{y}\geq\frac{(1+1)^2}{x+y}=\frac{4}{x+y}$$ and AM-GM
$$\frac{1}{xy}\geq\frac{1}{\left(\frac{x+y}{2}\right)^2}=\frac{4}{(x+y)^2}.$$
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