Is it true that $\forall x>3, \exists yx$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 ?$

Question

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!

Answer 1

It's easier if you substitute $a = 1/x, b=1/y, c=1/z$. I'll assume that we want $x,y,z > 0$. Then the conditions just say that $a,b,c$ are numbers between $0$ and $1$ so that $a + b + c = 1$ and $b$ is the biggest, $c$ is the smallest, and $a$ is in between. When is this possible?

You need $a < 1/2$ or else $b$ can't be bigger than $a$. But that's all you need: once you pick $a < 1/2$, just take $b$ to be $1/2$ and then take $c = 1 - a - b$.

Answer 2

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.

Answer 3

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

$\dfrac1x+\dfrac1y+\dfrac1z=\dfrac1x+1-\dfrac3{2x}+\dfrac1{2x}=1\;.$

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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

$\dfrac1x+\dfrac1y+\dfrac1z=\dfrac1x+\dfrac12+\dfrac12-\dfrac1x=1\;.$

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Hence we have already proved the following property :

Property :

$\forall x>2\,,\;\exists y<x\,$ and $\,\exists z>x\,$ such that

$\dfrac1x+\dfrac1y+\dfrac1z=1\;.$