Giúp mình với, cảm ơn nhiều luôn
View attachment 147775
áp dụng BĐT Cauchy có:
4, [tex]a^3+b^3+c^3\geq 3abc\geq abc (a;b;c\geq 0)[/tex]
5, [tex]+, \frac{a^6}{b^2}+\frac{a^6}{b^2}+\frac{a^6}{b^2}+\frac{b^6}{a^2}\geq 4\sqrt[4]{\frac{a^6}{b^2}.\frac{a^6}{b^2}.\frac{a^6}{b^2}.\frac{b^6}{a^2}}\geq 4a^4\\\\ +, \frac{b^6}{a^2}+\frac{b^6}{a^2}+\frac{b^6}{a^2}+\frac{a^6}{b^2}\geq 4b^4[/tex]
cộng các vế :>
6, [tex](a^5+b^5).(a+b)\geq (a^4+b^4).(a^2+b^2)\\\\ <=> a^6+ab^5+a^5b+b^6\geq a^6+a^2b^4+a^4b^2+b^6\\\\ <=> ab^5-a^2b^4+a^5b-a^4b^2\geq 0\\\\ <=> ab.(b^4-ab^3+a^4-a^3b)\geq 0\\\\ <=> b^3.(b-a)-a^3.(b-a)\geq 0 (vì ab>0)\\\\ <=> (b-a).(b-a).(b^2+ab+b^2)\geq 0 (đúng)[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
