Such that $x+y=1$, how can I show that for any positive real numbers $x,y$ we have $(1+\frac{1}{x})(1+\frac{1}{y})\geq 9$? Such that $x+y=1$, how can I show that for any positive real numbers $x,y$ we have $(1+\frac{1}{x})(1+\frac{1}{y})\geq 9$?
I see that $(1+\frac{1}{x})(1+\frac{1}{y})=1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}$ 
I really dont know how and from what I have to start with ? I know that  I have to use inequalities between means 
Can somone explain me the inequalitie's means and how to use them in problems with some exercises ? when I have to use GM or AM or QM or HM ? and what are the most popular simple basic inequalities that we have to use in problems  ? 
thank you :)
 A: $$(1+\frac{1}{x})(1+\frac{1}{y}) = (\frac{x+1}{x})(\frac{y+1}{y})=\frac{xy+x+y+1}{xy}=\frac{xy+2}{xy} = 1+\frac{2}{xy}$$
So we need to show that $xy \le \frac{1}{4}$ where $y=1-x$
A: Hint
Consider the function $$F=(1+\frac{1}{x})(1+\frac{1}{y})$$ and, from the constraint, replace $y$ by $(1-x)$. Now, look for the conditions where $F$ would be maximum or minimum ($F$ is just a function of $x$).
A: We know that $x + y = 1$, so we are interested in 
$$\left(1+\frac1x\right)\left(1 + \frac1y\right) = \left(1 + \frac1x\right)\left(1 + \frac1{1-x}\right) = \left(\frac{x+1}{x}\right)\left(\frac{2-x}{1-x}\right)$$
We are now concerned with finding the minimum of
$$\left(\frac{x+1}{x}\right)\left(\frac{2-x}{1-x}\right) = \frac{2 + x -x^2}{x(1-x)}$$
A: Cauchy-Schwarz gives
$$
\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\geq\left(1+\frac{1}{\sqrt{xy}}\right)^2\geq\left(1+\frac{1}{1/2}\right)^2=9
$$
where the the second inequality is because $\sqrt{xy}\leq\frac{1}{2}(x+y)=\frac{1}{2}$.
A: We have
$$
4\cdot \frac{1}{4} = 1\\
4\frac{(x + y)^2}{4} = 1\\
4 \left(\frac{x + y}{2}\right)^2 = 1\\
4\sqrt{xy}^2 \leq 1\\
4xy \leq 1\\
8xy \leq 2\\
8xy \leq 1 + x + y\\
9xy \leq 1 + x + y + xy\\
9 \leq \frac{1}{xy} + \frac1y + \frac1x + 1\\
9 \leq \left(1 + \frac1x\right)\left(1 + \frac1y\right)
$$
which is what we wanted to show. I used AM-GM from line 3 to line 4, I used that $x+y = 1$ from line 1 to line 2 as well as from line 6 to line 7.
A: \begin{align}(x+1)(y+1)\ge9xy & \impliedby (x+1)(2-x)\ge9x(1-x)\\&\impliedby 2x+2-x^2-x\ge9x-9x^2\\&\impliedby 8x^2-8x+2\ge0\\&\impliedby x^2-x+\dfrac{1}{4}\ge0\\&\impliedby \left(x-\dfrac{1}{2}\right)^2\ge0\end{align}
A: Another way would be to use Jensen inequality and convexity of $\log(1+\frac1t)$. 
