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

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 $u,v$
