Bài 2:

Ta cóB=\(\frac{8}{a^3b^3c^3}\)

Áp dụng BĐT cô-si, ta có 

\(a+b+c\ge3\sqrt[3]{abc}\Rightarrow3\sqrt[3]{abc}\le6\Rightarrow\sqrt[3]{abc}\le2\Rightarrow abc\le8\Rightarrow a^3b^3c^3\le512\)

=>\(\frac{8}{a^3b^3c^3}\ge\frac{8}{512}=\frac{1}{64}\)

dấu = xảy ra <=>a=b=c=2

^_^

Bài 1:,

Đặt ..=A

Ta có \(A=\frac{\left(c+a\right)\left(c+b\right)}{a+b}+\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(b+a\right)\left(b+c\right)}{a+c}\)

Đặt \(a+b=x;b+c=y;c+a=z\) =>x+y+z=2

Ta có A=\(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\)

Áp dụng BĐT cô-si, ta có \(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xy^2z}{zx}}=2y\)

Tương tự thì 2A\(\ge2\left(x+y+z\right)=4\Rightarrow A\ge2\left(ĐPCM\right)\)

Dấu = xảy ra <=>a=b=c=1/3

^_^

\(M=x^2+y^2+xy-3x-3y+2018\)

\(=x^2+2x\frac{\left(y-3\right)}{2}+\left(\frac{y-3}{2}\right)^2+y^2-3y+2018-\left(\frac{y-3}{2}\right)^2\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3y^2-6y+8063}{4}\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3\left(y^2-2y+1\right)}{4}+2015\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3\left(y-1\right)^2}{4}+2015\ge2015\)

\("="\Leftrightarrow x=y=1\)

Cảm ơn bạn nhiều nha

Thay x = y = 1 thì sẽ thấy nhé