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:
$$\frac{(x+y)(y+1)}{xy}\ge \frac{(x+y+2)^2}{(x+y)^2}$$
 A: Write $a=1/x>0$ and $b=1/y>0$. So we have to prove $$(1+a)(1+b)\geq (1+{2ab\over a+b})^2$$
or $$(a+b)^3+ab (a+b)^2\geq 4ab(a+b)+ 4a^2b^2$$
or $$(a+b)^2(a+b+ab)\geq 4ab(a+b+ ab)$$
or (since $a+b+ab>0$) $$(a+b)^2\geq 4ab$$ which is true.
A: It is equivalent to
$$ \frac{1}{2}\log\left(1+\frac{1}{x}\right)+\frac{1}{2}\log\left(1+\frac{1}{y}\right) \geq \log\left(1+\frac{1}{\frac{x+y}{2}}\right) $$
i.e. to the midpoint-convexity of $f(z)=\log\left(1+\frac{1}{z}\right)$ over $\mathbb{R}^+$. It simply follows from convexity, and convexity is a consequence of $f''(z)=\frac{1}{z^2}-\frac{1}{(z+1)^2}>0$.
A: 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
$$\frac{1}{xy}\geq\frac{1}{\left(\frac{x+y}{2}\right)^2}=\frac{4}{(x+y)^2}.$$
A: 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 $$(x+y)^2(x+y+1)\geq 4xy(x+y+1)$$
A: Since $\mathbb R^+×\mathbb R^+$ is convex, and the problem is symmetric in $x$ and $y$, the minimum of $$(1+\frac1x)(1+\frac1y)-(1+\frac2{x+y})^2 $$ occurs when $x=y$...
If $x=y$ we get $$(1+\frac1x)(1+\frac1x)-(1+\frac2{2x})^2=0$$...
It's a minimum since for instance  $(1,2)$ gives $3-\frac53=\frac43\gt 0$...
A: Why did nobody mention the one-liner solution: 
$$\left(1+\frac 1x\right)\left(1+\frac 1y\right)\stackrel{\text{Cauchy-Schwarz}}\geq\left(1+\frac {1}{\sqrt{xy}}\right)^2\stackrel{\text{AM-GM}}\geq\left(1+\frac{1}{\frac{x+y}{2}}\right)^2 = \left(1+\frac{2}{x+y}\right)^2.$$
A: Hint:$$\left (\frac{1}{x}-\frac{1}{y}\right )^2\geq 0$$
A: I give a proof using Cauchy-Schwartz for fun.
$$\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$$
$$\iff \Bigr((x+y)(1+\frac{1}{x})\Bigl)\Bigr((x+y)(1+\frac{1}{y})\Bigl)\ge \Bigr(x+y+2\Bigl)^2$$
$$\iff \Bigr(x+y+1+\frac{y}{x}\Bigl) \Bigr(x+y+1+\frac{x}{y}\Bigl) \ge \Bigr(x+y+2\Bigl)^2$$
$$\iff \Bigr(\sqrt x^2+\sqrt y^2+1^2+\sqrt{\frac{y}{x}}^2\Bigl) \Bigr(\sqrt x^2+\sqrt y^2+1^2+\sqrt{\frac{x}{y}}^2\Bigl) \\
\ge (\sqrt x \cdot \sqrt x + \sqrt y \cdot \sqrt y + 1\cdot1 + \sqrt{\frac{y}{x}} \cdot \sqrt{\frac{x}{y}})^2 $$
