Let $x$ and $y$ be two positive real numbers, such that $x + y = 1$. Prove that $$\left(1 +\frac{1}{x}\right)\left(1 +\frac{1}{y}\right) \ge 9$$
Problem
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Tags: algebra, inequalities
23.09.2021 19:17
23.09.2021 19:20
edit: sniped
23.09.2021 19:45
See also this post that was posted yesterday.
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
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 $x$.
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 $9$.
24.09.2021 04:01
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 $\frac{1}{x}+\frac{1}{y}\geq 4$
31.03.2024 20:37
(1+1/x)(1+1/y)=(2+y/x)(2+x/y)>=(2+1)²=9
31.03.2024 23:58
01.04.2024 01:53
AbhayAttarde01 wrote:
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
01.04.2024 03:40
Let $x$ and $y$ be two positive real numbers, such that $x + y = 1$. Prove that $$ \left(1 +\frac{1}{x}\right)\left(1 +\frac{1}{\sqrt y }\right)\ge 2(2+\sqrt 2)$$$$ \left(1 +\frac{1}{x}\right)\left(1 +\frac{\sqrt 2}{y}\right) \ge 2+3\sqrt 2+\sqrt {16+8\sqrt 2}$$