分式不等式a^2/b^2,abc/(a^3+b^3+c^3)

tommywong

teomihai:

If you have time for this
let a,b,c>0,prove that

$\displaystyle\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{9abc}{4(a^3+b^3+c^3)}\geq{\frac{15}{4}}$

thanks very much

tommywong

teomihai:

i solvi,t with this
Prove for any a,b,c>0

$\displaystyle\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right) \ge \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$
,but i don't prove that!, you have one hint?
thanks very much

kuing

回复 2# tommywong

这个我在《撸题集》P.948 记载过，与 1# 的题有啥关联吗？

tommywong

teomihai:

i solve this$\displaystyle\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right) \ge \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$

with that

$$(a+b+c)^2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-10(a^2+b^2+c^2)+(ab+bc+ca)=\sum \frac{(a-b)^2(b-2c)^2}{bc} $$
but is very ugly (how we find this decomposition?)
thanks very much

kuing

回复 4# tommywong

这个恒等式

青青子衿

回复 4# tommywong

这个恒等式怎么发现的？
样子有点奇怪……