# 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 :)

$$(1+\frac{1}{x})(1+\frac{1}{y}) = (\frac{x+1}{x})(\frac{y+1}{y})=\frac{xy+x+y+1}{xy}=\frac{xy+2}{xy} = 1+\frac{2}{xy}$$

So we need to show that $xy \le \frac{1}{4}$ where $y=1-x$

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 $x$).

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 $\sqrt{xy}\leq\frac{1}{2}(x+y)=\frac{1}{2}$.

• I always appreciate a good CS-application. Too bad I never seem to find them on my own. – Arthur Mar 13 '15 at 13:42

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 $$\left(\frac{x+1}{x}\right)\left(\frac{2-x}{1-x}\right) = \frac{2 + x -x^2}{x(1-x)}$$

• There's a typo $(x+1)(2-x)=-x^2+x+2$ . – Mr. Y Jan 5 '16 at 13:52
• @Mr.Y, fixed, thanks – jameselmore Jan 5 '16 at 14:44

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.

\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}

Another way would be to use Jensen inequality and convexity of $\log(1+\frac1t)$.