Giả sử c= max{a,b,c}
[tex]\Rightarrow \frac{(c-a)(c-b)}{abc+c^3}\geq 0[/tex]
Ta có
[tex]\frac{(a-b)(a-c)}{abc+a^3}+\frac{(b-c)(b-a)}{abc+b^3}\\\frac{a-b}{(abc+a^3)(abc+b^3)}.[(a-c)(abc+b^3)-(b-c)(abc+a^3)]\\=\frac{a-b}{(abc+a^3)(abc+b^3)}[abc(a-b)+c(a^3-b^3)]\\=\frac{a-b}{(abc+a^3)(abc+b^3)}.c[(a-b)(a+b)^2]\\=\frac{c(a-b)^2(a+b)^2}{(abc+a^3)(abc+b^3)}\geq 0[/tex]
Suy ra đpcm[/QUOTE]
harder & smarter là ai vậy -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
harder & smarter là ai vậy -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------