Quay lại hôm trước thì ta có đề bài : \(Cho ..a,b,c> 0 , x,y\geq 1.. thuoc R \\ CMR:\sqrt[3]{abc}\leq \sqrt[6]{\frac{[\sum (ab)^2+2x\sum a^2bc][\sum a^2+2y\sum ab]}{(3+6x)(3+6y) }}\leq \frac{1}{3}\sum a\)
Quay lại hôm nay thì ta có lời giải =)) tại đăng bài mà ko có sol thì tội lỗi :V 
\(\sum (ab)^2\geq \sum a^2bc,\sum a^2\geq ab(AM-GM)\\ \sqrt[6]{\frac{(\sum a^2b^2+2x\sum a^2bc)(\sum a^2+2y\sum ab)}{(3+6x)(3+6y)}}\geq \sqrt[6]{\frac{(1+2x)(1+2y)abc(a+b+c)\sum ab}{9(1+2x)(1+2y)}}=\sqrt[6]{\frac{\sum a\sum ab(abc)}{9}}\geq \sqrt[6]{(abc)^2}=\sqrt[3]{abc}(1)\\ f(x)=\sqrt[6]{\frac{(\sum a^2b^2+2x\sum a^2bc)(\sum a^2+2y\sum ab)}{(3+6x)(3+6y)}}(x\geq 1)\\ f'(x)=\sqrt[6]{\frac{\sum a^2+2y\sum ab}{3(3+6y)}}.\frac{2abc(a+b+c)(1+2x)-2[\sum a^2b^2+2xabc(a+b+c)]}{6(1+2x)^2.\sqrt[6]{\frac{\sum [(ab)^2+2xabc(a+b+c)]^5}{(1+2x)^5}}}\\ =\sqrt[6]{\frac{\sum a^2+2y\sum ab}{3(3+6y)}}.\frac{abc\sum a\sum (ab)^2}{3(1+2x)^2.\sqrt{\frac{[\sum (ab)^2+2xabc\sum a]^5}{(1+2x)^5}}}\\ => f(x)\leq f(1)\\CMTT:\sqrt[6]{\frac{(\sum a^2b^2+2\sum a^2bc)(\sum a^2+2y\sum ab)}{9(3+6y)}}\leq \sqrt[6]{\frac{(\sum a^2b^2+2\sum a^2bc)(\sum a^2+2y\sum ab)}{81}}=\sqrt[3]{\frac{\sum a\sum ab}{9}}.\\ Can cm \sqrt[3]{\frac{\sum a\sum ab}{9}}\leq \frac{\sum a}{3}=>(\sum a)^2\geq 3\sum ab(dung)...(2).Tu.(1)(2)=>ĐPCM\\ Dau''=''x=y=1,a=b=c\)