[不等式] a/b+b/c+c/a,28,12

$a$, $b$, $c>0$
\[\frac ab+\frac bc+\frac ca + \frac{28(ab+bc+ca)}{(a+b+c)^2} \geqslant 12.\]

貌似是道经典，不知这里有没有发过，晚上在不等式群看到，不知最早出自何处？

Proof:
Expanding and combining like terms,the inequality turns into
\[ \sum{\frac{a^3}{b}}+\sum{\frac{a^2b}{c}}+2\sum{\frac{ab^2}{c}}+7\sum{ab}\geq 10\sum{a^2} \]
or,equivalently,
\[ \left(\sum{\frac{a^3}{b}}+\sum{\frac{a^2b}{c}}-2\sum{\frac{ab^2}{c}}\right)+\left(4\sum{\frac{ab^2}{c}}-3\sum{ab}\right)\geq 10\sum{a^2}-\sum{ab} \]
Notice that:
\[ \left(\sum{\frac{a^3}{b}}+\sum{\frac{a^2b}{c}}-2\sum{\frac{ab^2}{c}}\right)=\sum{\frac{b(a-b)^{2}}{c}} \]
and
\[ \left(4\sum{\frac{ab^2}{c}}-3\sum{ab}\right)=\sum{\frac{c(2a-3b)^{2}}{b}} \]
Applying the AM-GM Inequality,we get
\[ \sum{\frac{b(a-b)^{2}}{c}} +\sum{\frac{c(2a-3b)^{2}}{b}}\geq 2\sum{(a-b)(2a-3b)} =10\sum{a^2}-10\sum{ab} \]
Hence we are done!

niubility的证法，不知是谁最先得到的？

3# kuing


当然是Can他们这群人了。另外，arqady提出
Let $a$, $b$ and $c$ are positive numbers. Prove that:
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+2\geq\frac{14(a^2+b^2+c^2)}{(a+b+c)^2}\]