Given $a, b,c,d$ non-negative and $a+b+c+d=4$. Prove that $a^3b+b^3c+c^3d+d^3a+23abcd \le 27.$ 
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Given $a, b,c,d$ are non-negative numbers and $a+b+c+d=4$. Prove that $$a^3b+b^3c+c^3d+d^3a+23abcd \le 27.$$


Given the problem, we use

$$
\begin{eqnarray}
a &=& u^2\\
b &=& v^2\\
c &=& w^2\\
d &=& x^2
\end{eqnarray} \tag 1
$$

So the boundary condition is given by

$$
u^2 + v^2 + w^2 + x^2 = 4, \tag 2
$$

for any $u$, $v$, $w$ and $x$.
We define

$$
S(u,v,w,x) = u^6 v^2 + v^6 w^2 + w^6 x^2 + x^6 u^2 + 23 u^2 v^2 w^2 x^2.
\tag 3
$$

Whence

$$
\begin{eqnarray}
\frac{\partial S}{\partial u} &=& 6 u^5 v^2 + 2 x^6 u + 46 u v^2 w^2 x^2,\\
\frac{\partial S}{\partial v} &=& 6 v^5 w^2 + 2 u^6 v + 46 u^2 v w^2 x^2,\\
\frac{\partial S}{\partial w} &=& 6 w^5 x^2 + 2 v^6 w + 46 u^2 v^2 w x^2,\\
\frac{\partial S}{\partial x} &=& 6 x^5 u^2 + 2 w^6 x + 46 u^2 v^2 w^2 x.\\
\end{eqnarray} \tag 4
$$


We know that

$$
u = v = w = x = 1 \tag 5
$$

satisfies the boundary condition and we obtain

$$
\frac{\partial S}{\partial u} = \frac{\partial S}{\partial v}
= \frac{\partial S}{\partial w} = \frac{\partial S}{\partial x} = 54. \tag 6
$$

But as

$$
dx = - du - dv - dw, \tag 7
$$

we obtain

$$
d S = \left( \frac{\partial S}{\partial u}
    - \frac{\partial S}{\partial x} \right) d u
    + \left( \frac{\partial S}{\partial v}
    - \frac{\partial S}{\partial x} \right) d v
    + \left( \frac{\partial S}{\partial w}
    - \frac{\partial S}{\partial x} \right) d w = 0. \tag 8
$$

So the "point" $(1,1,1,1)$ is an extreme value for $S$, and that value is $27$.

Another extreme "point " is $(0,0,0,2)$, and we obtain

$$
S(0,0,0,2) = 0 \tag 9
$$

So the extreme "point" $(1,1,1,1)$ is a maximum, whence

$$
u^6 v^2 + v^6 w^2 + w^6 x^2 + x^6 u^2 + 23 u^2 v^2 w^2 x^2 \le 27 \tag {10}
$$

Or going back to $(a,b,c,d)$, we get

$$
\bbox[16px,border:2px solid #800000] {
   a^3 b + b^3 c + c^3 d + d^3 a + 23 a b c d \le 27. } \tag {11}
$$

