Prove by induction that the equation $\dfrac{1}{x_1} + \dfrac{1}{x_2} + ... + \dfrac{1}{x_n} = 1$ has a solution with $x_1<x_2<...<x_n$.

Question

Using induction, I have to show that $\forall n \in \mathbb{N}, n \ge 3$ the equation:

$$\dfrac{1}{x_1} + \dfrac{1}{x_2} + ... + \dfrac{1}{x_n} = 1$$

has a solution $(x_1, x_2, ..., x_n) \in (\mathbb{N^*}, \mathbb{N^*}, ... \mathbb{N^*})$ (so a solution of all non-zero natural numbers) such that

$$x_1 < x_2 < ... < x_n$$

I don't even see how I could prove the base case. If we have:

$$p(3) : \dfrac{1}{x_1} + \dfrac{1}{x_2} + \dfrac{1}{x_3} = 1$$

I don't see the $3$ natural numbers that satisfy the equation and at the same time satisfy the condition $x_1 < x_2 < x_3$. So I can't even prove the base case, let alone imply $p(k + 1)$ if I assume $p(k)$ true. Even if I get the equation to a common denominator and multiply, I get:

$$x_1 x_2 + x_1x_3 + x_2 x_3 = x_1 x_2 x_3$$

So it didn't really get any more obvious. I thought that these might be the second and third Vieta's formulas for a $3$rd degree polynomial, but I don't see how I could construct a polynomial that has all $3$ of its roots natural and satisfy the condition $x_1 < x_2 < x_3$. Besides, the exercise clearly says that this should be proved using induction.

Answer 1

Start with $\frac12+\frac13+\frac16$. To increase $n$ by $1$, keep replacing the final fraction, say $\frac1m$, with $\frac{1}{m+1}+\frac{1}{m(m+1)}$.

Answer 2

Alternatively, start with $\dfrac12+\dfrac13+\dfrac16=1$

and go from $\dfrac{1}{x_1} + \dfrac{1}{x_2} + ... + \dfrac{1}{x_n} = 1$ to $\dfrac{1}{2x_1} + \dfrac{1}{2x_2} + ... + \dfrac{1}{2x_n}+\dfrac12 = 1$

Answer 3

Rewrite equation with $n+1$ unknow variables $$\dfrac{1}{x_1} + \dfrac{1}{x_2} + ... + \dfrac{1}{x_n} + \dfrac{1}{x_{n+1}}= 1$$ like this $$\dfrac{1}{x_2} + ... + \dfrac{1}{x_n}+\dfrac{1}{x_{n+1}} = \dfrac{x_{1}-1}{x_{1}}$$

so $$ \dfrac{1}{x_2 \dfrac{x_{1}-1}{x_{1}}} + ... + \dfrac{1}{x_n \dfrac{x_{1}-1}{x_{1}}} + \dfrac{1}{x_{n+1} \dfrac{x_{1}-1}{x_{1}}} =1$$

Let $a_1<a_2<...<a_n$ be solution (which by I.H. exists) to $$\dfrac{1}{a_1} + \dfrac{1}{a_2} + ... + \dfrac{1}{a_n} = 1$$ Let $x=x_{1}$. So we have to find $x_1,...x_{n+1}$ so that for each $k\geq 2$ we have $${x_k \dfrac{x-1}{x}} = a_{k-1}\implies a_{k-1}x=(x-1)x_k$$

If we put $x=$ then $x_k = 2a_{k-1}$. So $x_2<x_3<...$ and since clarly $a_1\geq 2$ we have $x_2 \geq 4>2=x_1$.