<s>[INST] Is it true that $\forall x>3, \exists yx$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 ?$ 
Is it true that $\forall x>3, \exists y<x$ and $\exists z>x$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 ?$

Here is my proof but I am stuck in the middle. And I am also wondering if I am on the right track?
Let $y = x - a$ for some real number $a$ that $a>0$, and $z = x + b$ for some real number $b$ that $b>0$.
Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 \Rightarrow \frac{1}{x}+\frac{1}{x-a}+\frac{1}{x+b}=1.$
After combining similar terms, I got:
$a = \frac{x(x^2+bx-3x-2b)}{x^2+(b-2)x-b}$, and $b = \frac{x(x^2-ax-3x+2a)}{-x^2+(a+2)x-a}$
For $a$ exist, $x^2+(b-2)x-b \neq 0$. And for $b$ exist, $-x^2+(a+2)x-a \neq 0$.
Let $f(x) = x^2+(b-2)x-b $  and  $g(x) = -x^2+(a+2)x-a $.
By the formula of quadratic function: $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$:
$f(x)\neq 0$ $\Rightarrow (b-2)^2+4b < 0$ ,and $g(x)\neq 0\Rightarrow(a+2)^2-4a<0$.
Since $(b-2)^2+4b \geq 0$,and $(a+2)^2-4a\geq0$ , no matter how we choose $a$ and $b$,  $f(x)=0$ and $g(x)=0$ always have a solution.
Then I stuck at here, I understand that I need to prove $f(x)\neq0$ and $g(x)\neq0$ for all $x >3$ and $a,b$ have to be positive to make the statement true. Otherwise, it's false. Really hope to get some hints/help!! Thank you!
 [/INST] 
Is it true that $\;\forall x>3\,,\;\exists y<x\,$ and $\,\exists z>x\,$ such that $\,\dfrac1x+\dfrac1y+\dfrac1z=1\;?$

Yes, it is true.
Proof :
First of all it results that
$\dfrac x{x-\frac32}<\dfrac x{3-\frac32}=\dfrac23x<x\;\;$ for any $\,x>3\,.$
Consequently, for all $\,x>3\,$ there exist $\,y=\dfrac x{x-\frac32}<x\,$ and $\,z=2x>x\,$ such that
$\dfrac1x+\dfrac1y+\dfrac1z=\dfrac1x+1-\dfrac3{2x}+\dfrac1{2x}=1\;.$

Addendum :
$\forall x\in\big(2,3\big]\,,\;\exists y<x\,$ and $\,\exists z>x\,$ such that
$\dfrac1x+\dfrac1y+\dfrac1z=1\;.$
Proof :
For any $\,x\in\big(2,3\big]\,$ it results that
$\dfrac{2x}{x-2}>\dfrac{2x-x(4-x)}{x-2}=\dfrac{x^2-2x}{x-2}=x\;.$
Consequently, for all $\,x\in\big(2,3\big]\,$ there exist $\,y=2<x\,$ and $\,z=\dfrac{2x}{x-2}>x\,$ such that
$\dfrac1x+\dfrac1y+\dfrac1z=\dfrac1x+\dfrac12+\dfrac12-\dfrac1x=1\;.$

Hence we have already proved the following property :
Property :
$\forall x>2\,,\;\exists y<x\,$ and $\,\exists z>x\,$ such that
$\dfrac1x+\dfrac1y+\dfrac1z=1\;.$
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