It's a rare situation
when the nature of equation
makes an honest computation
beat a smart consideration,
when you shouldn't contemplate,
only differentiate...
In other words that is a case when to show that the function is increasing, the best thing to do is just to compute the derivative and show that it is positive.
I prefer to write the integral as $\iiint |F|^2\log(|F|^2)$ where $F(u,v,w)=au+bv+cw$, $a,b,c>0$ and $u,v,w$ run over the unit circle independently. We will fix one variable and show that for the fixed sum of squares of the coefficients at the other two, the closer they are, the larger the integral is. Due to rotation invariance, it suffices to show that for every fixed $x>0$,
$$
I(a,b,x)=\iint G\overline G\log(G\overline G)
$$
where $G(u,v)=x+au+bv$, $a,b>0$ is an increasing function of $a$ in the regime $a^2+b^2=\operatorname{const}, a<b$.
Taking the derivative and using that $ada+bdb=0$, i.e., $(da,db)$ is proportional to $(b,-a)$, we get the inequality
$$
\Re\iint (bu-av)(x+a\bar u+b\bar v)(2\log|x+au+bv|+1)\ge 0
$$
to prove.
The product of the first parentheses equals
$x(bu-av)+(b^2u\bar v-a^2v\bar u$, so $1$ in the third factor integrates to $0$ and we are left with
$$
b\Re\int\left[(x+b\bar v)\int u\log|x+au+bv|\right]\ge
a\Re\int\left[(x+a\bar u)\int v\log|x+au+bv|\right]\,.
$$
Now we just note that for every $z\in \mathbb C$, we have
$$
\int u\log|z+u|=\pi z^*
$$
where $z^*=\begin{cases}z& |z|\le1\\ \frac{1}{\bar z} & |z|\ge 1\end{cases}$.
So we need to prove that
$$
b\int(x+b\bar v)\left(\frac{x+bv}{a}\right)^*\ge
a\int(x+a\bar u)\left(\frac{x+au}{b}\right)^*\,.
$$
Now it is clear that the integrands are real positive, so $\Re$ can be dropped.
If $|x+bv|>a$ for all $v$, the first integral evaluates to $ba$ and the second does not exceed $ab$, so the inequality is trivial. Otherwise we can make a triangle with sides $a,b,x$. Let $\alpha$ and $\beta$ be the angles opposite to $a$ and $b$ respectively. Another honest computation
of the integrals in question reduces the task to proving that
$$
(\pi-\alpha)ab+b\frac{x^2+b^2}a\alpha-\frac{2xb^2}a\sin\alpha\ge
(\pi-\beta)ba+a\frac{x^2+a^2}b\beta-\frac{2xa^2}b\sin\beta\,.
$$
Taking into account that the sides of the triangle are proportional to the sines of the opposite angles, we reduce this to
$$
\alpha\left[-\sin\alpha\sin\beta+\sin\beta\frac{\sin^2(\alpha+\beta)+\sin^2\beta}{\sin\alpha}\right]-2\sin^2\beta\sin(\alpha+\beta)\ge
\\
\text{the same expression with $\alpha$ and $\beta$ swapped.}
$$
We can rewrite the LHS as
$$
\frac{\sin\beta}{\sin\alpha}
\left[\alpha(\sin^2(\alpha+\beta)+\sin^2\beta-\sin^2\alpha)-
2\sin\alpha\sin\beta\sin(\alpha+\beta)\right]\,.
$$
Now we use the trigonometric identity
$$
\sin^2(\alpha+\beta)+\sin^2\beta-\sin^2\alpha=2\sin(\alpha+\beta)\sin\beta\cos\alpha
$$
and, cancelling the common factors, obtain
$$
\frac{\sin^2\beta}{\sin\alpha}[\alpha\cos\alpha-\sin\alpha]
\ge \frac{\sin^2\alpha}{\sin\beta}[\beta\cos\beta-\sin\beta]\,,
$$
so it suffices to prove that $\alpha\mapsto\frac{\sin\alpha-\alpha\cos\alpha}{\sin^3\alpha}$ is increasing on $(0,\frac\pi 2)$.
Taking the derivative, getting rid of the denominator, and using $\sin^2+\cos^2=1$, we reduce it to
$$
\alpha+2\alpha\cos^2\alpha>3\cos\alpha\sin\alpha\,.
$$
Passing to $\psi=2\alpha\in(0,\pi)$, we rewrite it as
$$
2\psi+\psi\cos\psi\ge 3\sin\psi\,.
$$
At $0$ we have equality, so we take the derivative again and want
$$
2-2\cos\psi-\psi\sin\psi>0\,.
$$
Again equality at $0$, so again the derivative:
$$
\sin\psi-\psi\cos\psi>0\,.
$$
Now one could think a bit at last, but the inertia of "computing honestly like a horse", as one of my friends puts it, is too great, so we have equality at $0$, take the derivative, and get
$$
\psi\sin\psi>0.
$$
Here one should stop the mindless computations and finally observe that both factors are positive on $(0,\pi)$, so the inequality is trivial.
The End.
