2011年的圣诞节，越南Le Hai贴出了下面不等式并祝福所有Mathlinker圣诞快乐！

Let $a,\;b,\;c$ be reals such that $a+b+c=0$ and $a^2+b^2+c^2=3.$ Prove that \[{a^5}b + {b^5}c + {c^5}a \le  - 3.\]
 

事实上，用Maple因式分解不难得到它等价于

\[ (a^3+6a^2b+3ab^2-b^3)^2\geq0 \]

只要做点齐次化工作就好。

但过了很久都没看见更理想的证明，直到前不久，tian275461网友给了一个基于三角函数的证明，Can神表示他有Cauchy-Schwarz proof，但他就是不肯说。

Proof (tian 275461)

we use $c=-a-b$put in $a^2+b^2+c^2=3$then  $a^2+ab+b^2=\dfrac{3}{2}$
let $a=x-y,b=x+y,\Longrightarrow c=-2x  $,and $3x^2+y^2=\dfrac{3}{2}$
so we can let
$x=\dfrac{\sqrt{2}}{2}\cos{t},y=\dfrac{\sqrt{6}}{2}\sin{t},t\in(0,2\pi) $
\[\begin{align*}
a^5+b^5c+c^5a&=(x-y)^5(x+y)-2x(x+y)^5-32x^5(x-y)\\
&=\left(\dfrac{\sqrt{2}}{2}\cos{t}-\dfrac{\sqrt{6}}{2}\sin{t}\right)^5\left(\dfrac{\sqrt{2}}{2}\cos{t}+\dfrac{\sqrt{6}}{2}\sin{t}\right)\\
&-2\dfrac{\sqrt{2}}{2}\cos{t}\left(\dfrac{\sqrt{2}}{2}\cos{t}+\dfrac{\sqrt{6}}{2}\sin{t}\right)^5\\
&-32\left(\dfrac{\sqrt{2}}{2}cost\right)^5\left(\dfrac{\sqrt{2}}{2}\cos{t}-\dfrac{\sqrt{6}}{2}\sin{t}\right)\\
&=-\dfrac{3}{8}\left[10-2\sin{\left(6t-\dfrac{\pi}{6}\right)}\right]\le -3
\end{align*} \]

同时，下面推广可以同样证明：

1 if $a+b+c=0,a^2+b^2+c^2=3$,then

\[-\dfrac{3\sqrt{26}}{4}\le a^4b+b^4c+c^4a\le\dfrac{3\sqrt{26}}{4}\]

2 if $a+b+c=0,a^2+b^2+c^2=3$,then
\[-\dfrac{3\sqrt{2}}{2}\le a^2b+b^2c+c^2a\le\dfrac{3\sqrt{2}}{2}\]

3: if $a+b+c=0,a^2+b^2+c^2=3$,then
\[-\dfrac{9\sqrt{38}}{8}\le a^6b+b^6c+c^6a\le\dfrac{9\sqrt{38}}{8}\]

希望以后能看见Can神的CS proof.

。。。。。

2013年3月1日，Vo Quoc Ba Can 终于发了究极的Cauchy-Schwarz proof
Proof. Since $a+b+c=0$ and $a^2+b^2+c^2=3,$ it is easy to obtain the below results:

\[\left\{
  \begin{array}{ll}
    ab+bc+ca=-\dfrac{3}{2}, &  \\
   a^3b+b^3c+c^3a=-(ab+bc+ca)^2=-\dfrac{9}{4}, &\\
   ab^2+bc^2+ca^2+3abc=-(a^2b+b^2c+c^2a), &  \\
   a^3b^3+b^3c^3+c^3a^3=(ab+bc+ca)^3+3a^2b^2c^2=-\dfrac{27}{8}+3a^2b^2c^2, &  \\
   \displaystyle\sum (4ab+2c^2+6bc+3)^2=54, &
  \end{array}
\right.\]

With these results, we have
\begin{align*} {a^5}b + {b^5}c + {c^5}a& = \sum {{a^5}b}\\
 &= \sum {{a^3}b(3 - {b^2} - {c^2})} \\
 &= 3\sum {{a^3}b} - \sum {{a^3}{b^3}} - abc\sum {a{b^2}} \\
  & = - \frac{{27}}{4} + \frac{{27}}{8} - 3{a^2}{b^2}{c^2} - abc\sum {a{b^2}}\\
  &= - \frac{{27}}{8} + abc\sum {{a^2}b} .
\end{align*}
 Therefore, it suffices to prove that
 \begin{equation}
  abc(a^2b+b^2c+c^2a) \le \frac{3}{8}.
 \end{equation}
On the other hand, using the Cauchy-Schwarz inequality, we have
 \[{\left[ {\sum {a(4ab + 2{c^2} + 6bc + 3)} } \right]^2} \le \left( {\sum {{a^2}} } \right)\left[ {\sum {{{(4ab + 2{c^2} + 6bc + 3)}^2}} } \right] = 162.\]
  From this, it follows that
  \[-9\sqrt{2} \le \sum a(4ab+2c^2+6bc+3) \le 9\sqrt{2},\]
   or
   \[-\frac{3}{\sqrt{2}} \le a^2b+b^2c+c^2a+3abc \le \frac{3}{\sqrt{2}}.\]
    The last inequality yields:
    \begin{equation}
      (a^2b+b^2c+c^2a+3abc)^2 \le \frac{9}{2}
    \end{equation}
 Using (2) and the AM-GM inequality, we have
\begin{align*}
 abc(a^2b+b^2c+c^2a) &=\frac{1}{3}\cdot 3abc\cdot (a^2b+b^2c+c^2a)\\ & \le \frac{1}{3} \left(\frac{3abc+a^2b+b^2c+c^2a}{2}\right)^2\le \frac{3}{8},
\end{align*}
      which is (1). So, we are done.