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let $xneq y$ 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: $(sqrt{x}-sqrt{y})^2geq0$ , 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 ?

Following the hints of the comments, your condition implies that
$$sqrt{x} + sqrt{y} = 1$$
AM-GM states
$$1 geq 2sqrt{sqrt{xy}} implies 1/4 geq sqrt{xy}$$
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 $sqrt{xy}$,
$$geq (1+ 4)^2 =25$$

Just to give a different approach, let $x=1/u^2$ and $y=1/v^2$ with $u,vgt0$. Then

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

provided $unot=v$ (i.e., provided $xnot=y$), which allows the cancellation of the $v-u$ term. Note this now requires $u,vgt1$. We also have

$${v+uover uv}=1implies uv=u+vimplies u^2v^2=(v+u)^2=u^2+v^2+2uv=u^2+v^2+2u+2v$$

It follows that

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

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

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

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