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!

For every  $n\in N \gt 2$, we show that there exists $n$ pairwise  distinct natural numbers whose sum of inverses equals $1$.
Proof by Induction.
Making induction on $n\geq 3$, the initial case $n=3$ following from
$\frac {1}{2} + \frac {1}{3}+\frac {1}{6}=1$.
We now assume by Induction hypothesis, that, for a certain natural number $k\geq 3 $ there exists natural numbers  $x_1\lt x_2\lt ...\lt x_k$ such that
$\frac {1}{x_1} + \frac {1}{x_2}+...+\frac {1}{x_k}=1$.
Multiplying through  the equality above by $\frac {1}{2}$ and adding $\frac {1}{2}$ on both sides of the resulting equality , we obtain
$\frac {1}{2}+\frac {1}{2x_1} + \frac {1}{2x_2}+...+\frac {1}{2x_k}=1$.
Since  $2\lt 2x_1\lt 2x_2\lt ...\lt 2x_k$, we have $k+1$ pairwise distinct natural numbers whose sum of inverses equals  $1$ and this completes the induction step.
