Question

Cho a,b,c là các số thực không âm thỏa mãn:a+2b+3c=4.CMR:\(\left(a^2b+b^2c+c^2a+abc\right)\left(ab^2+bc^2+ca^2+abc\right)\)≤8

Answer 1

\(8VT=4\left(a^2b+b^2c+c^2a+abc\right)\left(2ab^2+2bc^2+2ca^2+2abc\right)\le\left(a^2b+b^2c+c^2a+2ab^2+2bc^2+2ca^2+3abc\right)^2\)

\(\Rightarrow VT\le\frac{1}{32}\left(2a^2b+2b^2c+2c^2a+4ca^2+4ab^2+4bc^2+6abc\right)^2\)

\(\Rightarrow VT\le\frac{1}{32}\left(2a^2b+2b^2c+2c^2a+4ca^2+4ab^2+4bc^2+9abc\right)^2\)

\(\Rightarrow VT\le\frac{1}{32}\left[\left(a+2b\right)\left(b+2c\right)\left(c+2a\right)\right]^2\)

\(\Rightarrow VT\le\frac{1}{512}\left[\left(a+2b\right)\left(4b+8c\right)\left(c+2a\right)\right]^2\)

\(\Rightarrow VT\le\frac{1}{512}\left(\frac{a+2b+4b+8c+c+2a}{3}\right)^6=\frac{1}{512}\left(a+2b+3c\right)^6=\frac{4^6}{512}=8\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;1;0\right)\)