<s>[INST] Given $a, b,c,d$ non-negative and $a+b+c+d=4$. Prove that $a^3b+b^3c+c^3d+d^3a+23abcd \le 27.$ 
This is a problem on Mathlinks.ro and it have had no solution. So I hope it would have a nice answer and as simply as possible.
Given $a, b,c,d$ are non-negative numbers and $a+b+c+d=4$. Prove that $$a^3b+b^3c+c^3d+d^3a+23abcd \le 27.$$

 [/INST] We can trade off the cyclic symmetry to reduce the number of variables to consider.  The homogeneous form is:
$$4^4(a^3b+b^3c+c^3d+d^3a+23abcd) \le 27(a+b+c+d)^4$$
WLOG, let $a$ be the minimum among $a, b, c, d$; and so we have non-negatives $x, y, z$ s.t. $b = a+x, \, c = a+y, \, d = a+z$.  Hence we get an inequality of form
$$f(x, y, z)\cdot a^2+g(x, y, z)\cdot a+h(x, y, z) \ge 0$$
It is easy to check $f = 32\cdot23[(x-y)^2+(y-z)^2+(z-x)^2] + 32\cdot11(x^2+y^2+z^2) \ge 0$  
$g = 16 (11x^3 + 33x^2y + 81x^2z + 81xy^2 - 206xyz + 81xz^2 + 11y^3 + 33y^2z + 81yz^2 + 11z^3)$ which is non-negative as AM-GM gives:
$$11x^3 + 33x^2y + 81x^2z + 81xy^2 + 81xz^2 + 11y^3 + 33y^2z + 81yz^2 + 11z^3 \\ \ge 9\cdot9\cdot11^{5/9} \cdot xyz \ge 9\cdot9\cdot3\cdot xyz \ge 206xyz$$
It only remains to show $h \ge 0$, which is the same as setting one variable to zero, or showing $4^4(a^3b+b^3c) \le 27(a+b+c)^4$.  
In fact it is not hard to prove the stronger cyclic version $4^4(a^3b+b^3c+c^3a) \le 27(a+b+c)^4$ or the tougher version $a+b+c=4 \implies a^3b+b^3c+c^3a+\frac{473}{64}abc \le 27$ using exactly the same technique to reduce variables from $3 \to 2$!
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