Prove $abc+abd+acd+bcd\le\frac{1}{27}+\frac{176abcd}{27}$ for $a+b+c+d=1$ 
Let $a,b,c$, and $d$ be four positive reals satisfying $a+b+c+d=1$. Show that
$$abc+abd+acd+bcd\le\frac{1}{27}+\frac{176abcd}{27}.$$

I tried the inequality between $27abc$ and $(a+b+c)^3$ but it didn't help me
 A: Inspired by https://math.stackexchange.com/a/1748125/823641.

Define $f(a,b,c,d) = abc+abd+acd+bcd - \frac{1}{27} - \frac{176abcd}{27}$.
We have $f(1/4,1/4,1/4,1/4) = 0$.
We shall show that $f(1/4,1/4,1/4,1/4)$ is the maximum among all $a,b,c,d$ satisfying the conditions.
Suppose not, and assume that another value of $(a,b,c,d) = (x,y,z,w) \neq (1/4,1/4,1/4,1/4)$ corresponds to the maximum $f(x,y,z,w)$.
WLOG, we assume that $x \neq y$.
We have
\begin{align}
&f(\frac{x+y}{2},\frac{x+y}{2},z,w) - f(x,y,z,w) \\
=~&(x-y)^2(z+w)/4 - \frac{176zw}{27 * 4}(x-y)^2 \\
=~& \frac{(x-y)^2}{4}(z+w-\frac{176zw}{27}),
\end{align}
where
$$\frac{(x-y)^2}{4} > 0.$$
If $z+w-\frac{176zw}{27} > 0$ then we are done because in that case we have $f(\frac{x+y}{2},\frac{x+y}{2},z,w) > f(x,y,z,w)$, completing the proof by contradicting the assumption of maximality.
Now, we assume that
$z+w-\frac{176zw}{27} \leq 0$,
then we have
$$f(x,y,z,w) = xy(z+w - 176zw/27) + (x+y)zw - 1/27$$
$$\leq (x+y)zw - 1/27 \leq (1 - t)t^2/4 - 1/27,$$
where we have used the Cauchy–Schwarz inequality and $t = z+w$.
Let $g(t) = (1 - t)t^2/4 - 1/27$.
It is easy to check (by $g'(t)$) that $g(t)$ achieves its maximum at $t = 2/3$ with $g(t) = 0$.
Therefore, we have $f(x,y,z,w) \leq 0$, completing the proof.
A: We rewrite the inequality as :
$$f(a,b,c,d)=-\left(\ln\left(a\right)+\ln\left(b\right)+\ln\left(c\right)+\ln\left(d\right)+\ln\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\frac{176}{27}\right)\right)\geq -\ln\left(1/27\right)$$
We use the equal variable method corollary 1.5 as :
$$a+b+c+d=1$$
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=constant$$
$$f\left(x\right)=-\ln\left(x\right)$$ define :
$$g\left(x\right)=f'\left(\frac{1}{\sqrt{x}}\right)$$
then we have :
$$g''(x)>0$$
So the $f(a,b,c,d)$ is minimal for $d\geq c\geq b \geq a>0$ and $d=c=b=x$
Remains to show the inequality for $x\in[0,1/3)$ :
$$x^{3}\left(1-3x\right)\left(\frac{3}{x}+\frac{1}{1-3x}-\frac{176}{27}\right)-1/27\leq 0$$
Or :
$$1/27(3x-1)(4x-1)^{2}(11x+1)\leq 0$$
Can you end now ?


Adding remark for another proof :
For $0<a\leq b\leq c \leq d$, $a+b+c+d=1$ and $b=c=0.25$ we have the inequalities :
$$\left(abcd\left(\frac{1}{a}+\frac{9\left(1+\frac{5\left(a-0.25\right)^{2}}{1+a}\right)}{1-a}\right)-\frac{1}{27}-\frac{176}{27}abcd\right)\geq abcd\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)-\frac{1}{27}-\frac{176}{27}abcd$$
Wich is a one variable inequality .
Now and again fixing $b=c=0.25$ and introducing the variable $u,v>0$ such that $a+u+v+d=1=a+b+c+d$ and $0<a\leq b\leq c \leq d$ we have :
$$abcd\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)-\frac{1}{27}-\frac{176}{27}abcd\geq auvd\left(\frac{1}{a}+\frac{1}{u}+\frac{1}{v}+\frac{1}{d}\right)-\frac{1}{27}-\frac{176}{27}auvd$$
using the previous inequality .
Now it misses one equality case so to be continued...
Edit 01/09/2022 :
We have the following inequalities for $0<a<0.3\leq b \leq c \leq d$ and $a+b+c+d=1$ :
$$0\geq \left(abcd+\frac{a^{2}\left(0.25-a\right)^{2}}{2\left(2-4a\right)}\right)\left(\frac{1}{a}+\frac{9}{\left(d+c+b\right)}-\frac{176}{27}\right)-\frac{1}{27}\geq abcd\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\frac{176}{27}\right)-\frac{1}{27}$$
And we can use the same strategy fixing two new variable $u,v$
