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

Question

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 $$a^3b+b^3c+c^3d+d^3a+23abcd \le 27.$$

Answer 1

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 $3 \to 2$!

Answer 2

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

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

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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 $27$.

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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

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