Let $f(x) = \dfrac{1}{2x^2-6x+9}$. First let us first quickly sketch some properties of $f$ that we'll need. $f(x)=\dfrac{1}{2(x-\frac32)^2+\frac92}$, so its maximum is $f(\frac32)=\frac29$. $f~'(x)=\dfrac{6-4x}{(2x^2-6x+9)^2}$, and $f~''(x)=12\dfrac{2x^2-6x+3}{(2x^2-6x+9)^3}$. Setting $f~''(x)=0$ gives the inflection points $x=\frac32\pm\frac{\sqrt3}2$. $f$ is increasing before $\frac32$, decreasing after $\frac32$, and is concave between the inflection points. Note $f(\frac32\pm\frac{\sqrt3}2)=\frac16$.
If $a$, $b$, and $c$ are all between the inflection points, then by Jensen's inequality
$$f(a)+f(b)+f(c) \le 3f(\dfrac{a+b+c}3) = 3f(1) = \frac35$$
as required.
If at least two of $a$, $b$, or $c$ are outside the inflection points, then
$$f(a)+f(b)+f(c) \le \frac16+\frac16+\frac29=\frac59\lt\frac35$$
If one of $a$, $b$, or $c$ is $\gt\frac32+\frac{\sqrt3}2$, then at least one of $a$, $b$, or $c$ is $\lt\frac32-\frac{\sqrt3}2$, otherwise
$$3=a+b+c\ge 2(\frac32-\frac{\sqrt3}2)+(\frac32+\frac{\sqrt3}2)=\frac92-\frac{\sqrt3}2 \approx 3.7,$$
a contradiction. And so by the previous paragraph, the required result holds.
So the only remaining case is when exactly one of $a$, $b$, or $c$ is $\lt\frac32-\frac{\sqrt3}2$, and the other two are between the inflection points. We need two estimates. First consider the case when the minimum is $\le\frac12$. (Here $\frac12$ is arbitrary, although we need to choose something very close to $\frac12$ for the following argument to work.) Then
$$f(a)+f(b)+f(c) \le f(\frac12) + \frac29 + \frac29 = \frac{2}{13} + \frac49 = \frac{70}{117} \lt \frac35$$
Finally, suppose $a$ is between $\frac12$ and $\frac32-\frac{\sqrt3}2$, and $b$ and $c$ are between the inflection points. By Jensen's inequality, we need only consider the case when $b=c$, by replacing them with their mean. Then $b=(3-a)/2$. So $a\gt\frac12$ implies $b\lt\frac54$, and
$$f(a)+f(b)+f(b) \lt f(\frac32-\frac{\sqrt3}2) + 2f(\frac54) = \frac16 + \frac{16}{37} = \frac{133}{222} \lt \frac35$$