The statement is exactly the one in https://mathworld.wolfram.com/PrasolovPoint.html - in OP notations, the ortic triangle is $\Delta H_aH_bH_c$, and its reflection in the nine-point circle is $\Delta N_aN_bN_c$.
For a structural proof, the simplest way to go is to compute points using barycentric coordinates.
This is not only showing the required property, but also computes the barycentric coordinates of the point of concurrence in question, from the formula we can see its complexity, recover the result claimed in
ETC,
and further exhibit interesting other properties - like seeing it is on the line $X(5)X(6)$.
I am giving the human solution, complemented for the calculus with computer algebra assistance, easy digestible code, which may be useful for some readers (in this and similar situations).
Let $a,b,c$ be the lengths of the sides in $\Delta ABC$. A point $P$ has normed barycentric coordinates $(x,y,z)$ if $P=xA+yB+zC$,
with the norming condition $1=x+y+z$. (The relation can be considered using affixes, or vectorially for some / any chosen origin point $\Omega$ in the affine plane so that the vectors relation holds $\Omega P=x\;\Omega A +y\; \Omega B + z\;\Omega C$.)
Sometimes it is better to use "projective" coordinates $[x:y:z]$ instead, with $x+y+z\ne 0$, and we pass to the corresponding normed coordinates by dividing by the sum $s=x+y+z$. We often write $\frac 1s(x,y,z)$ instead of $(x/s,\ y/s,\ z/s)$ for this normed version.
Let $S$ being the area of $\Delta ABC$, so $S^2 =s(s-a)(s-b)(s-c)$. The barycentric coordinates of the nine-point center $X(5)$ are $[f(a,b,c)\ :\ f(b,c,a)\ :\ f(c,a,b)]$, see ETC, where
$$
f(a,b,c)=a^2(b^2 + c^2) - (b^2 -c^2)^2\ .
$$
The norming factor for this representation of $X(5)$ is
$$
\begin{aligned}
&\sum f(a,b,c):=f(a,b,c) + f(b,c,a)+f(c,a,b)
\\
&\qquad
=\sum a^2(b^2 + c^2) - (b^2 -c^2)^2
=\sum a^2b^2 + a^2c^2 + 2b^2 c^2 - b^4 - c^4
\\
&\qquad=4a^2b^2 + 4b^2c^2 + 4 c^2 a^2 - 2a^4 - 2b^4 - 2c^4
\\
&\qquad=2(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=32s(s-a)(s-b)(s-c)
\\
&\qquad=32S^2\ .
\end{aligned}
$$
Our aim is to show explicitly:
Proposition: There exists an intersection point $X(68)=AN_a\cap BN_b\cap CN_c$, and its barycentric coordinates are
$$
\bbox[yellow]{\qquad
X(68)=
[ \ g(a,b,c)\ :\ g(b,c,a)\ :\ g(c,a,b)\ ]
\qquad}
$$
with
$$
\bbox[yellow]{\qquad
g(a,b,c) =
\frac{-a^2+b^2+c^2}{a^4 + b^4 + c^4 - 2a^2(b^2+c^2))}\ .
\qquad}
$$
Bonus: The point $X(68)$ also lies on $X(5)X(6)$, the line through the nine-point center $X(5)$, and the symmedian point $X(6)=[a^2:b^2:c^2]$.
Proof: We compute:
$$
\begin{aligned}
A &= (1,0,0)\ ,\\[2mm]
\frac{H_aC}{BC} &=\frac 1a\cdot b\cos C=\frac 1a\cdot b\cdot \frac{a^2 + b^2 -c^2}{2ab}
=\frac 1{2a^2}(a^2+b^2-c^2)\ ,\\
H_a&=\left(0,\frac{H_aC}{BC},\frac{BH_a}{BC}\right)=
\frac 1{2a^2}(0,\ a^2 + b^2 -c^2,\ a^2 - b^2 + c^2)\ ,
\\[2mm]
X(5) &= (\ x_{X(5)},\ y_{X(5)},\ z_{X(5)}\ )\\
&=
\phantom{\frac 1{8S^2}}
[\
a^2(b^2+c^2) - (b^2-c^2)^2\ :\
b^2(a^2+c^2) - (a^2-c^2)^2\ :\
c^2(a^2+b^2) - (a^2-b^2)^2\ ]\ ,\\
&=
\frac 1{32S^2}(\
a^2(b^2+c^2) - (b^2-c^2)^2\ ,\
b^2(a^2+c^2) - (a^2-c^2)^2\ ,\
c^2(a^2+b^2) - (a^2-b^2)^2\ )\ ,
\\[2mm]
N_a &= 2X(5)-H_a=(\ x_{Na},\ y_{Na},\ z_{Na}\ )\ ,
\\[2mm]
y_{Na}
&=2\cdot \frac 1{32S^2}\cdot(b^2(a^2+c^2) - (a^2-c^2)^2)
-\frac 1{2a^2}\cdot (a^2 + b^2 -c^2)
\\
&=\frac 1{32a^2S^2}\Big(\ 2a^2b^2(a^2+c^2)-2a^2(a^2-c^2)^2 - 16S^2(a^2 + b^2 -c^2)\ \Big)\ .
\\[2mm]
32a^2S^2\; y_{Na}
&=2a^2b^2(a^2+c^2)-2a^2(a^2-c^2)^2 - 16S^2(a^2 + b^2 -c^2)
\\
&=2a^2b^2(a^2-b^2+c^2)+2a^2 b^4-2a^2(a^2-c^2)^2 - 16S^2(a^2 + b^2 -c^2)
\\
&=2a^2b^2(a^2-b^2+c^2) + 2a^2 (a^2 + b^2 -c^2)(-a^2+b^2+c^2)
- 16S^2(-a^2 + b^2 -c^2) - 32a^2S^2
\\
&=(a^2-b^2+c^2)(2a^2b^2 +16S^2) + 2a^2 (a^2 + b^2 -c^2)(-a^2+b^2+c^2)
- 32a^2S^2
\\[2mm]
&\qquad
(a^2 + b^2 -c^2)(-a^2+b^2+c^2) - 16S^2
\\
&\qquad\qquad=(b^4 - a^4 - c^4 + 2a^2c^2) - (-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2)
\\
&\qquad\qquad=-2b^2(a^2-b^2+c^2)\ ,
\\[2mm]
32a^2S^2\; y_{Na}
&=(a^2-b^2+c^2)(2a^2b^2 +16S^2)-4a^2b^2(a^2-b^2+c^2)
\\
&=(a^2-b^2+c^2)(-2a^2b^2+16S^2)
\\
&=(a^2-b^2+c^2)(-a^4-b^4-c^4+2a^2c^2+2b^2c^2)\ .
\\
32a^2S^2\; z_{Na}
&=(a^2+b^2-c^2)(-a^4-b^4-c^4+2a^2b^2+2b^2c^2)\text{ similarly .}
\end{aligned}
$$
The line $AN_a$ has the equation in barycentric coordinates for a point $P=(x,y,z)$ given
by assembly of the coordinates for $A,N_a$, and $P$ in matrix format, than building the determinant:
$$
0=
\begin{vmatrix}
1 & 0 & 0\\
x_{N_a} & y_{Na} & z_{N_a}\\
x & y & z
\end{vmatrix}\ ,
$$
so the equation involves only $y,z$ with constant proportionality:
$$
\begin{aligned}
\frac yz
&=
\frac{y_{Na}}{z_{Na}}
=
\frac
{(a^2-b^2+c^2)(-a^4-b^4-c^4+2a^2c^2+2b^2c^2)}
{(a^2+b^2-c^2)(-a^4-b^4-c^4+2a^2b^2+2b^2c^2)}
\\
&=
\frac
{(a^2-b^2+c^2):(-a^4-b^4-c^4+2a^2b^2+2b^2c^2)}
{(a^2+b^2-c^2):(-a^4-b^4-c^4+2a^2c^2+2b^2c^2)}
\\
&=\frac{g(b,c,a)}{g(c,a,b)}
\ .
\end{aligned}
$$
Similar equations are written for $BN_b$, and $CN_c$, and the proof is finished by observing that the point $X(68)$ from the proposition satisfies the above equation of $AN_a$, $y:z=g(b,c,a):g(c,a,b)$, as by symmetriy it also the case for $BN_b$, and $CN_c$.
$\square$
Sage code:
var('a,b,c');
Ha = 1/2/a^2 * vector([0, a^2 + b^2 - c^2, a^2 - b^2 + c^2])
f(a, b, c) = a^2*(b^2 + c^2) - (b^2 - c^2)^2
X5 = vector([f(a, b, c), f(b, c, a), f(c, a, b)])
X5 = 1/sum(X5) * X5 # thus norming it
Na = 2*X5 - Ha
xNa, yNa, zNa = Na
Then:
print(factor(yNa / zNa))
delivers immediately the result of our longer computation:
(a^4 + b^4 - 2*a^2*c^2 - 2*b^2*c^2 + c^4)*(a^2 - b^2 + c^2)
/
((a^4 - 2*a^2*b^2 + b^4 - 2*b^2*c^2 + c^4)*(a^2 + b^2 - c^2))
(Result was manually rearranged to fit in the MSE line.)
We can now also show the bonus point, by computing the determinant:
with lines proportional to the barycentric coordinates of respectively
$X(6)$, $X(5)$, and $X(68)$, and showing it vanishes:
$$
\begin{vmatrix}
a^2 & b^2 & c^2\\
f(a,b,c) & f(b,c,a) & f(c,a,b)\\
g(a,b,c) & g(b,c,a) & g(c,a,b)
\end{vmatrix}
$$
g(a,b,c) = (-a^2 + b^2 + c^2) / (a^4 + b^4 + c^4 - 2*a^2*(b^2 + c^2))
matrix([ [a^2, b^2, c^2],
[f(a, b, c), f(b, c, a), f(c, a, b)],
[g(a, b, c), g(b, c, a), g(c, a, b)], ]).det().simplify_full()
And the code delivers rapidly the zero.
Best Answer
The right inequality.
In the standard notation we need to prove that: $$\sum_{cyc}\frac{a^2\cdot\frac{2bc\cos\frac{\alpha}{2}}{b+c}}{\frac{2ac\cos\frac{\beta}{2}}{a+c}\cdot\frac{2ab\cos\frac{\gamma}{2}}{a+b}}\geq\frac{2(a+b+c)}{\sqrt3}$$ or $$\sum_{cyc}\frac{a^2\cdot\frac{2bc\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}}{b+c}}{\frac{2ac\sqrt{\frac{1+\frac{a^2+c^2-b^2}{2ac}}{2}}}{a+c}\cdot\frac{2ab\sqrt{\frac{1+\frac{a^2+b^2-c^2}{2ab}}{2}}}{a+b}}\geq\frac{2(a+b+c)}{\sqrt3}$$ or $$\sum_{cyc}\frac{\frac{a^2\sqrt{bc(a+b+c)(b+c-a)}}{b+c}}{\frac{\sqrt{ac(a+b+c)(a+c-b)}}{a+c}\cdot\frac{\sqrt{ab(a+b+c)(a+b-c)}}{a+b}}\geq\frac{2(a+b+c)}{\sqrt3}$$ or $$\sum_{cyc}\frac{a(a+b)(a+c)}{b+c}\sqrt{\frac{b+c-a}{(a+b-c)(a+c-a)}}\geq\frac{2\sqrt{(a+b+c)^3}}{\sqrt3}$$ or$$\sum_{cyc}\frac{a(a+b)(a+c)(b+c-a)}{b+c}\geq2\sqrt{\frac{(a+b+c)^3\prod\limits_{cyc}(a+b-c)}{3}}.$$ Now, let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that: $$\sum_{cyc}\frac{x(y+z)(2y+x+z)(2z+x+y)}{2x+y+z}\geq8\sqrt{\frac{(x+y+z)^3xyz}{3}}.$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$, and $xyz=w^3$.
Thus, we need to prove that: $$\sum_{cyc}\frac{(3v^2-yz)(3u+y)^2(3u+z)^2}{\prod\limits_{cyc}(3u+x)}\geq24\sqrt{u^3w^3}$$ or $$\frac{-w^6+171u^3w^3-9uv^2w^3+405u^4v^2-54u^2v^4}{w^3+54u^3+9uv^2}\geq8\sqrt{u^3w^3}$$ or $f(v^2)\geq0,$ where $$f(v^2)=-w^6+171u^3w^3-9uv^2w^3+405u^4v^2-54u^2v^4-8(w^3+54u^3+9uv^2)\sqrt{u^3w^3}.$$ But since by Maclaurin $$u\geq v\geq w,$$ we obtain: $$f'(v^2)=405u^4-108u^2v^2-9uw^3-72u^2\sqrt{uw^3}>0,$$ which says that $f$ increases.
Thus, it's enough to prove our inequality for a minimal value of $v^2$, which by $uvw$ happens for equality case of two variables.
Since our inequality is homogeneous, it's enough to assume that $y=z=1,$ which gives $$\frac{2x(x+3)^2}{2x+2}+\frac{2(x+1)(x+3)(2x+2)}{x+3}\geq8\sqrt{\frac{(x+2)^3x}{3}}$$ or $$x(x+3)^2+4(x+1)^3\geq8(x+1)\sqrt{\frac{(x+2)^3x}{3}},$$ which after squaring of the both sides gives $$(x-1)^2(11x^4+50x^3+91x^2+88x+48)\geq0$$ and we are done!