Let $x$ and $y$ be two positive real numbers, such that $x + y = 1$. Prove that $$\left(1 +\frac{1}{x}\right)\left(1 +\frac{1}{y}\right) \ge 9$$
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Tags: algebra, inequalities
Bole
23.09.2021 19:17
Expand to get \[1+\frac{x+y+1}{xy} \geq 9\]this simplifies to \[\frac{2}{xy} \geq 8\]From AM-GM we know that \[(x+y)^2 \geq 4xy\]or \[\frac{1}{4} \geq xy\]Therefore \[\frac{2}{xy} \geq 8\]and equality is achieved at \[x=y=\frac 12\]
Andrew_Dou
23.09.2021 19:20
Substitute $x=a-1$ and $y=b-1,$ then it follows that $a+b=3.$ Note that by AM-GM, we have that $$\frac{9}4 \geqslant ab \implies 27 \geqslant 8ab+9 \implies ab+9a+9b \geqslant 9ab+9 \implies ab \geqslant 9(a-1)(b-1) \implies (\frac{a}{a-1})(\frac{b}{b-1}) \geqslant 9 \implies (1+\frac{1}x)(1+\frac{1}y) \geqslant 9,$$where the second to last step is possible since $a,b \neq 1.$
edit: sniped
programjames1
23.09.2021 19:45
By AM-GM,
$$\left(1+\frac1x\right)\left(1+\frac1y\right) = \left(1+\frac{x}{x}+\frac{y}{x}\right)\left(1+\frac{x}{y}+\frac{y}{y}\right)\ge \left(3\sqrt[3]{\frac{y}{x}}\right)\left(3\sqrt[3]{\frac{x}{y}}\right) = 9.$$
Consider the function
$$f(x) = \ln\left(1+\frac1x\right) = \ln(x+1) - \ln(x).$$Notice that
$$f''(x) = \frac{1}{x^2} - \frac{1}{(x+1)^2} > 0$$for all $x > 0.$ This means $f$ is concave, so Jensen's inequality gives
$$f(x)+f(y)\ge 2f\left(\frac{x+y}{2}\right)$$$$\Longleftrightarrow$$$$\left(1+\frac1x\right)\left(1+\frac1y\right)\ge e^{2\ln 3} = 9.$$
See also this post that was posted yesterday.
SK1729
23.09.2021 19:58
Using AM Gm inq (x+y)/2 ge sqrt (xy) or xy is le 1/4 now on expanding the given lhs of the inq we get 1+(x+y+1)/xy=1 +2/xy which is ge 1+2/(1/4)=1+8=9 hence proved equality hold at xy=1/2
Grolarbear
23.09.2021 21:30
Factoring out, we have $1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \ge 9$. $xy + x + y + 1 \ge 9xy$ $(x+1)(y+1) \ge 9xy$ We have $x = 1-y$. So, $-y^2+y+2 \ge -9y^2+9y$. $8y^2 -8y+2 \ge 0$. The root of this quadratic is at $(\frac{1}{2}, 0$ Hence, $y \ge 0$ which is true). By the same method/symmetrically, this is the same fo $x$.
OofPirate
23.09.2021 21:31
$x+y=1$. We can see that $x=1-y$ and $y=1-x$. Apply the $AM-GM$ inequality; $(1+\dfrac{1}{x})(1+\dfrac{1}{y})=(1+\dfrac{y}{x}+\dfrac{x}{x})(1+\dfrac{y}{y}+\dfrac{x}{y})\geq (3\sqrt[3]{\dfrac{y}{x}})(3\sqrt[3]{\dfrac{x}{y}})$ in which It is equal to $9$.
OlympusHero
24.09.2021 04:01
Expanding gives $1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy} = 1 + \frac{x+y+1}{xy} \geq 9$, so it suffices to show that $\frac{2}{xy} \geq 8$. But AM-GM gives $xy \leq \frac{1}{4}$, so $\frac{2}{xy} \geq 8$ as desired.
JPEG
24.09.2021 04:12
Expand gives $1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}$, using Am-Gm we have $xy \leq \frac{(x+y)^{2}}{4}=\frac{1}{4}$ and $\frac{1}{x}+\frac{1}{y}\geq 4$
Gvegve
31.03.2024 20:37
(1+1/x)(1+1/y)=(2+y/x)(2+x/y)>=(2+1)²=9
AbhayAttarde01
31.03.2024 23:58
the lowest value will be when the values of $x$ and $y$ are equal to each other, giving us that $x=y=0.5$, for the smallest case. We substitute and solve this equation, giving us $3*3 = 9$, which is greater than or equal to $9$, which shows that any other values other than $0.5$ for both will result in a value higher than $9$.
Q.E.D
anduran
01.04.2024 01:53
AbhayAttarde01 wrote:
the lowest value will be when the values of $x$ and $y$ are equal to each other, giving us that $x=y=0.5$, for the smallest case. We substitute and solve this equation, giving us $3*3 = 9$, which is greater than or equal to $9$, which shows that any other values other than $0.5$ for both will result in a value higher than $9$.
Q.E.D
But how can you prove that minimal value is attained at $x=y=\frac{1}{2}$?
sqing
01.04.2024 03:24
parmenides51 wrote: Let $x$ and $y$ be two positive real numbers, such that $x + y = 1$. Prove that $$\left(1 +\frac{1}{x}\right)\left(1 +\frac{1}{y}\right) \ge 9$$ https://artofproblemsolving.com/community/c6h2583852p22296948
sqing
01.04.2024 03:40
Let $x$ and $y$ be two positive real numbers, such that $x + y = 1$. Prove that $$ \left(1 +\frac{1}{x}\right)\left(1 +\frac{1}{\sqrt y }\right)\ge 2(2+\sqrt 2)$$$$ \left(1 +\frac{1}{x}\right)\left(1 +\frac{\sqrt 2}{y}\right) \ge 2+3\sqrt 2+\sqrt {16+8\sqrt 2}$$