We can rewrite the given equation as:

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}$

Multiplying both sides by $abc(a+b+c)$, we get:

$bc(a+b+c) + ac(a+b+c) + ab(a+b+c) = abc$

Expanding, we have:

$ab^2 + ac^2 + a^2b + a^2c + b^2c + bc^2 + abc + acb + bca = abc$

Rearranging terms, we get:

$a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 + 2abc = abc$

Subtracting $abc$ from both sides, we have:

$a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 + abc = 0$

Now, let's consider the expression $1/a^{2019} + 1/b^{2019} + 1/c^{2019} - 1/a^{2019} - 1/b^{2019} - 1/c^{2019}$.

Using the formula for the difference of cubes, we can factor this expression as:

$\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)\left(\frac{1}{a^{2018}} - \frac{1}{b^{2018}}\right)\left(a^{2018} - b^{2018}\right)$

Since $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}$, the first factor is equal to 1.

Also, $a^{2018} - b^{2018}$ can be factored as $(a-b)(a^{2017} + a^{2016}b + \ldots + ab^{2016} + b^{2017})$.

Therefore, the expression becomes:

$\left(\frac{1}{a^{2018}} - \frac{1}{b^{2018}}\right)(a-b)(a^{2017} + a^{2016}b + \ldots + ab^{2016} + b^{2017})$

Since $a-b$ is a factor, we can cancel it out from both sides.

This leaves us with:

$\left(\frac{1}{a^{2018}} - \frac{1}{b^{2018}}\right)(a^{2017} + a^{2016}b + \ldots + ab^{2016} + b^{2017}) = 0$

Since $a^{2017} + a^{2016}b + \ldots + ab^{2016} + b^{2017}$ is a sum of terms, it cannot be equal to zero.

Therefore, the only way for the expression to be equal to zero is if $\frac{1}{a^{2018}} - \frac{1}{b^{2018}} = 0$.

This implies $\frac{1}{a^{2018}} = \frac{1}{b^{2018}}$, which means $a^{2018} = b^{2018}$.

Similarly, we can show that $b^{2018} = c^{2018}$.

Therefore, $a^{2018} = b^{2018} = c^{2018}$.

Taking the 2019th power of both sides, we get:

$a^{2018 \cdot 2019} = b^{2018 \cdot 2019} = c^{2018 \cdot 2019}$

Therefore, $a^{2019} = b^{2019} = c^{2019}$.

Substituting this into the original equation, we have:

$\frac{1}{a^{2019}} + \frac{1}{b^{2019}} + \frac{1}{c^{2019}} = \frac{1}{a^{2019}} + \frac{1}{b^{2019}} + \frac{1}{c^{2019}}$

Therefore, $1/a^{2019} + 1/b^{2019} + 1/c^{2019} = 1/a^{2019} + 1/b^{2019} + 1/c^{2019}$.

Hence, the statement is true.