Any 'odd unit fraction' whose denominator is not $1$ can be represented as the sum of three different 'odd unit fractions'? Let us call a fraction whose denominator is odd 'odd fraction'. Also, let us call an odd fraction whose numerator is 1  'odd unit fraction'.
Then, here is my question.

Question : Is the following true?
"Any odd unit fraction whose denominator is not $1$ can be represented as the sum of three different odd unit fractions."

Motivation : I've been asking this question. Then, I reached the above expectation.
Examples :
$$\frac 13=\frac 15+\frac 19+\frac 1{45}$$
$$\frac 15=\frac 1{7}+\frac 1{21}+\frac 1{105}$$
$$\frac 17=\frac 19+\frac 1{33}+\frac 1{693}$$
$$\frac 19=\frac 1{11}+\frac 1{51}+\frac 1{1683}$$
$$\vdots$$
$$\frac 1{99}=\frac 1{101}+\frac 1{5001}+\frac 1{16668333}$$
$$\vdots$$

If $n$ is not a multiple of 3, then $$\frac{1}{n}=\frac{1}{n+2}+\frac{3}{(n+2)(n+4)}+\frac{1}{n(n+2)(n+4)}$$
Second try: You have the first fraction is $1/(n+2)$.  Then you want to solve $$\frac{2}{n(n+2)}=\frac{1}{a}+\frac{1}{b}$$
Rearrange that into $(2a-n(n+2))(2b-n(n+2))=n^2(n+2)^2$  You can make the right-hand side equal to $pq$, where both $p$ and $q$ are $3\bmod 4$.  (Take $p=n$ or $p=n+2$) Then $a$ and $b$ will be odd.
So $a=n(n+3)/2, b=n(n+2)(n+3)/2$ or $a=(n+1)(n+2)/2, b=n(n+2)(n+1)/2$
