Câu hỏi của dbrby

Question

cho a,b,c > 0 thỏa mãn a+b+c=6abc.

Cmr: $\frac{bc}{a^3\left(c+2b\right)}+\frac{ca}{b^3\left(a+2c\right)}+\frac{ab}{c^3\left(b+2a\right)}\ge2$

Nguyễn Việt Lâm · Accepted Answer

$a+b+c=6abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6$

Đặt $\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=6$

$P=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}=\frac{x^4}{xy+2zx}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}$

$P\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\frac{\left(xy+yz+zx\right)^2}{3\left(xy+yz+zx\right)}=\frac{xy+yz+zx}{3}=2$

Dấu "=" xảy ra khi $x=y=z=\sqrt{2}$ hay $a=b=c=\frac{1}{\sqrt{2}}$Câu trả lời: $a+b+c=6abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6$

Đặt $\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=6$

$P=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}=\frac{x^4}{xy+2zx}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}$

$P\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\frac{\left(xy+yz+zx\right)^2}{3\left(xy+yz+zx\right)}=\frac{xy+yz+zx}{3}=2$

Dấu "=" xảy ra khi $x=y=z=\sqrt{2}$ hay $a=b=c=\frac{1}{\sqrt{2}}$