if $x-y =\sqrt{x}-\sqrt{y}$ with $x\neq y$ then $(1+\frac{1}{x})(1+\frac{1}{y})\geq 25$?

Question

let $x\neq y$ be positive real numbers such that :$x-y= \sqrt{x}-\sqrt{y}$ , I have tried to prove this inequality $(1+\frac{1}{x})(1+\frac{1}{y})\geq 25$ that i have created but i didn't got it.

Attempt I have showed that:$(\frac{1}{x}+\frac{1}{y})\geq \frac{2}{\sqrt{xy}}$ using this identity: $(\sqrt{x}-\sqrt{y})^2\geq0$ , I also showed that :$\frac{1}{xy}\geq \frac{1}{16}$ , Now I have used both result I have got the following inequality :$(1+\frac{y}{x})(1+\frac{x}{y})\geq 25$ but not what i have claimed , any way ?

Answer 1

Following the hints of the comments, your condition implies that $$\sqrt{x} + \sqrt{y} = 1$$ AM-GM states $$1 \geq 2\sqrt{\sqrt{xy}} \implies 1/4 \geq \sqrt{xy}$$ Use Cauchy on $$(1+1/x)(1+1/y) \geq (1+1/\sqrt{xy})^2$$ Pluggin in what you got in the AM-GM for $\sqrt{xy}$, $$\geq (1+ 4)^2 =25$$

Answer 2

Just to give a different approach, let $x=1/u^2$ and $y=1/v^2$ with $u,v\gt0$. Then

$$\begin{align} x-\sqrt x=y-\sqrt y &\implies{1\over u^2}-{1\over v^2}={1\over u}-{1\over v}\\ &\implies{(v-u)(v+u)\over u^2v^2}={v-u\over uv}\\ &\implies{v+u\over uv}=1\\ &\implies{1\over u}+{1\over v}=1 \end{align}$$

provided $u\not=v$ (i.e., provided $x\not=y$), which allows the cancellation of the $v-u$ term. Note this now requires $u,v\gt1$. We also have

$${v+u\over uv}=1\implies uv=u+v\implies u^2v^2=(v+u)^2=u^2+v^2+2uv=u^2+v^2+2u+2v$$

It follows that

$$\begin{align} \left(1+{1\over x}\right)\left(1+{1\over y}\right) &=(1+u^2)(1+v^2)\\ &=1+u^2+v^2+u^2v^2\\ &=1+2u^2+2v^2+2u+2v\\ &={(2u+1)^2+(2v+1)^2\over2}\\ &\gt{(2\cdot1+1)^2+(2\cdot1+1)^2\over2}\\ &=25 \end{align}$$

Answer 3

By AM-GM $$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\left(1+\frac{4}{4x}\right)\left(1+\frac{4}{4y}\right)\geq$$ $$\geq\frac{5}{\sqrt[5]{(4x)^4}}\cdot\frac{5}{\sqrt[5]{(4y)^4}}=\frac{25}{\sqrt[5]{4^{8}\left(\sqrt{xy}\right)^8}}\geq\frac{5}{\sqrt[5]{4^8\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^{16}}}=25.$$

Answer 4

The function $$y\mapsto \log\left(1+\tfrac{1}{y^2}\right)$$ is convex on $(0,\infty)$, so Jensen's inequality $$\mathrm{E}\left[\,\log\left(1+\tfrac{1}{Y^2}\right)\right]\geq \log\left(1+\tfrac{1}{\mathrm{E}[Y]^2}\right)$$ holds for any positive random variable $Y$. Letting $Y=\sqrt{X}$, in particular we have $$\mathrm{E}\left[\,\log\left(1+\tfrac{1}{X}\right)\right]\geq \log\left(1+\tfrac{1}{\mathrm{E}[\sqrt{X}]^2}\right)$$ for any positive random variable $X$.

Now, let $X$ be a categorical variable taking on the $N$ positive values $x_0,x_1,\ldots,x_{N-1}$ with equal probability $\tfrac{1}{N}$, and suppose that $$\mathrm{E}[\sqrt{X}]=\frac{1}{N}\sum_{i=0}^{N-1}\sqrt{x_i}=\frac{1}{N}\text{.}$$ Then the inequality above is

$$\frac{1}{N}\sum_{i=0}^{N-1}\log\left(1+\tfrac{1}{x_i}\right)\geq \log(1+N^2)\text{;}$$ taking exponentials, we have shown that $$\boxed{\sum_{i=0}^{N-1}\sqrt{x_i}=1\Rightarrow\prod_{i=0}^{N-1}\left(1+\tfrac{1}{x_i}\right)\geq (1+N^2)^N }\text{.}$$ OP's inequality is the case $N=2$.