[不等式] Le Hai 不等式

Let $a,\,b,\,c$ be real numbers satisfying $a+b+c=0$ and $a^2+b^2+c^2=3.$ Prove that \[a^5b+b^5c+c^5a \le -3.\]
来看看Can的神柯西。
Proof. Since $a+b+c=0$ and $a^2+b^2+c^2=3,$ it is easy to obtain the below results:
\begin{align*}
ab+bc+ca&=-\frac{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.\\
\sum (4ab+2c^2+6bc+3)^2&=54.\\  
\end{align*}

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
\[abc(a^2b+b^2c+c^2a) \le \frac{3}{8}. \qquad  (1) \]
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: \[(a^2b+b^2c+c^2a+3abc)^2 \le \frac{9}{2}. \qquad (2)\]
    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.

果然如标题所示: 厉害不等式

2# kuing


戳中笑点了。哈哈。。。