Note that $$\left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right) = \frac{(x+1)(y+1)}{xy} = \frac{xy + x+y+1}{xy} = \frac{xy+2}{xy} = 1 + \frac{2}{xy} = (*).$$Now, by AM-GM, $$x+y \geq 2\sqrt{xy}\iff \frac{1}{2}\geq \sqrt{xy}\iff \frac{1}{4}\geq xy\iff 4 \leq \frac{1}{xy}.$$Thus, $(*) \geq 1 + 2\cdot 4 = 9$, as desired. Equality holds when $x = y = \frac{1}{2}$.

A simpler alternative: $1+1/x = (x+1)/x = (2x+y)/x$. Hence, it suffices showing $(2x+y)(x+2y) \ge 9xy$. Now, $2x+y = x+x+y\ge 3\sqrt[3]{x^2y}$. Likewise, $x+2y\ge 3\sqrt[3]{xy^2}$. Multiplying these concludes the proof.

Even simpler: $\left(1+\frac {1}{x}\right)\left(1+\frac {1}{y}\right)=\left(2+\frac {y}{x}\right)\left(2+\frac {x}{y}\right) \ge 3^2$ by C-S.

$\prod_{\text{cyc}}\left(1+\frac1x\right)=\frac{xy+2}{xy}=1+\frac2{xy}\ge1+\frac2{\frac{(x+y)^2}4}=9$ Where the last inequality is by GM-AM in the denominator.

The problem it is to prove that x+y+1>=8xy or xy<=1/4, but this is just AM-GM

Expand and simplify with $x+y=1$: \begin{align*} \left(1+\frac 1x\right)\left(1+\frac 1y\right) &\ge 9\\ 1+\frac 1x+\frac 1y+\frac{1}{xy} &\ge 9\\ \frac{x+y}{xy}+\frac{1}{xy} &\ge 8\\ \frac{1}{xy} &\ge 4\\ \frac14 &\ge xy.\end{align*}Now make the substitution $y=1-x$ to get $$\frac14\ge x(1-x)=x-x^2 \iff x^2-x+\frac14\ge 0.$$Now, set $f(x)=x^2-x+\frac14$. Trivially, $f'(x)=2x-x$ so $f'(x)=0$ at $x=\frac12$. Therefore, the minimum of $f(x)=x^2-x+\frac14$ is $f\left(\frac12\right)=0$, so we can conclude that $$f(x)=x^2-x+\frac14\ge 0$$which finishes the problem. Also, someone please fix the latex Thanks @below

It was the parentheses (the \right was outside of the parens) franzliszt wrote: \begin{align*} \left(1+\frac 1x\right)\left(1+\frac 1y\right)&\ge 9\\ 1+\frac 1x+\frac 1y+\frac{1}{xy} &\ge 9\\ \frac{x+y}{xy}+\frac{1}{xy} &\ge 8\\ \frac{1}{xy} &\ge 4\\ \frac14 &\ge xy.\end{align*}

Put $x=sin \theta$ and $y=cos\theta$ After that result follows easily

Com10atorics wrote: Let $x,y$ be positive real numbers such that $x+y=1$. Prove that $\left(1+\frac {1}{x}\right)\left(1+\frac {1}{y}\right)\geq 9$. 1971 Canada 1992 Denmark https://artofproblemsolving.com/community/c4h1127976p5223590: Let $x,y>0$, $x+y\leq 1$. Prove that: $$(1-\frac{1}{x^2})(1-\frac{1}{y^2})\geq 9$$$$(1-\frac{1}{x^3})(1-\frac{1}{y^3})\geq 49$$$$ (1-\frac{1}{x^2})(1+\frac{1}{ y^3}) \leq -27$$https://artofproblemsolving.com/community/c6h3242887p29737427

Observe that $(1+\frac{1}{x})(1+\frac{1}{y})\geq 9 $ $\Leftrightarrow$ $(x+y)+xy+1\geq 9xy$ $\Leftrightarrow$ $1\geq 4xy$ $\Leftrightarrow$ $(x+y)^2\geq 4xy$ $\Leftrightarrow$ $(x-y)^2\geq 0 $

Let $ x,y>0 $ and $x+y\le 4$. Prove that $$\left(1+\frac {1}{x}\right)\left(1+\frac {1}{y}\right)\geq \frac {9}{4}$$$$\left(1+\frac {1}{x}\right)\left(1+\frac {2}{y}\right)\geq \frac {8+\sqrt{15}}{4}$$

sqing wrote: Let $ x,y>0 $ and $x+y\le 4$. Prove that $$\left(1+\frac {1}{x}\right)\left(1+\frac {1}{y}\right)\geq \frac {9}{4}$$Let $ x,y>0 $ and $x+y\le 3+\sqrt{15}$. Prove that $$\left(1+\frac {1}{x}\right)\left(1+\frac {1}{y}\right)\geq \frac {5}{3}$$ Statement. Let $x,y,k>0$ and $x+y\leq k$. Prove that $$\left ( 1+\frac 1x\right )\left (1+\frac 1y\right )\geq \frac{(k+2)^2}{k^2}$$Proof. From AM-GM we get $xy\leq \left(\frac{x+y}{2}\right)^2\leq \frac{k^2}{4}$. Then we have $$\left ( 1+\frac 1x\right )\left (1+\frac 1y\right )= \frac{(x+1)(y+1)}{xy}$$$$(\text{C-S}) \geq \frac{(\sqrt {xy}+1)^2}{xy}=1+\frac{2}{\sqrt {xy}}+\frac{1}{xy}\geq \frac{(k+2)^2}{k^2}.$$The equality holds when $x=y=\frac k2$. $\square$

https://artofproblemsolving.com/community/c6h3349512p31064865

Let $ x,y>0 $ and $x+y\le 3 $. Prove that $$\left(1+\frac {1}{x}\right)\left(1+\frac {3}{y}\right)\geq 3+ \frac {4\sqrt 2}{3}$$$$\left(1+\frac {1}{x}\right)\left(1+\frac {1}{3y}\right)\geq \frac {41+4\sqrt{10}}{27}$$

franzliszt wrote: Expand and simplify with $x+y=1$: \begin{align*} \left(1+\frac 1x\right)\left(1+\frac 1y\right) &\ge 9\\ 1+\frac 1x+\frac 1y+\frac{1}{xy} &\ge 9\\ \frac{x+y}{xy}+\frac{1}{xy} &\ge 8\\ \frac{1}{xy} &\ge 4\\ \frac14 &\ge xy.\end{align*}Now make the substitution $y=1-x$ to get $$\frac14\ge x(1-x)=x-x^2 \iff x^2-x+\frac14\ge 0.$$Now, set $f(x)=x^2-x+\frac14$. Trivially, $f'(x)=2x-x$ so $f'(x)=0$ at $x=\frac12$. Therefore, the minimum of $f(x)=x^2-x+\frac14$ is $f\left(\frac12\right)=0$, so we can conclude that $$f(x)=x^2-x+\frac14\ge 0$$which finishes the problem. Also, someone please fix the latex Thanks @below good

Using GMHM: $$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\geq \left(\frac{2}{\frac{x}{2x+y}+\frac{y}{2y+x}}\right)^2$$And thus it remains to prove $$\frac{x}{2x+y}+\frac{y}{2y+x}\leq \frac{2}{3}$$Which is equivalent to $$\left(1-\frac{x+y}{2x+y}\right) +\left(1-\frac{x+y}{2y+x}\right) \leq \frac{2}{3}$$Or $$(x+y)\left(\frac{1}{2x+y}+\frac{1}{x+2y}\right) \geq \frac{4}{3}$$Which is trivial by C-S.

Same as vsamc, just compressed the AM-GM part: The only inequality we use is AM-GM on $xy$ in the denominator: \begin{align*} \left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)&=\frac{(x+1)(y+1)}{xy} = \frac{xy+x+y+1}{xy}=\frac{xy+2}{xy}=1+\frac{2}{xy}\\ &\ge 1+\frac{2}{\left(\frac{x+y}{2}\right)^2}=1+8=9, \end{align*}and we are done.