Chủ đề này có 2 trả lời

[Peteroldar]

Cho $a, b, c$ là các số thực dương tuỳ ý. Chứng minh rằng:

$(a+b+c)(ab+bc+ca)\leq \frac{8}{9}(a+b)(b+c)(c+a)$

 

[DOTOANNANG]

$9(\,a+ b\,)(\,b+ c\,)(\,c+ a\,)- 8(\,a+ b+ c\,)(\,ab+ bc+ ca\,)=$ $(\,a^{\,2}b+ b^{\,2}c+ c^{\,2}a- 3\,abc\,)$ $+ (\,ab^{\,2}+ bc^{\,2}+ ca^{\,2}- 3\,abc\,)$

[DOTOANNANG]

$$8(\,a^{\,2}+ b^{\,2}+ c^{\,2}\,)(\,a+ b+ c\,)- 9(\,a+ b\,)(\,b+ c\,)(\,c+ a\,)\geqq 0$$

Spoiler

$<$$=$$>$ $(\,4\,a+ 4\,b+ 3\,c)(\,a- b\,)^{\,2}+ (\,4\,b+ 4\,c+ 3\,a)(\,b- c\,)^{\,2}+ (\,4\,c+ 4\,a+ 3\,b)(\,c- a\,)^{\,2}\geqq 0$

Mạnh hơn$:$ $\lceil$ https://diendantoanh...đt/#entry716367 $\rfloor$

$<$$=$$>$ $(\,a^{\,2}+ b^{\,2}+ c^{\,2}+ 7\,ab+ 7\,bc+ 7\,ca\,)(\,a+ b+ c\,)\geqq 9(\,a+ b\,)(\,b+ c\,)(\,c+ a\,)$