Question

giúp mk với cần gấp

cho a,b,c>0 tm 1/a +1/b+1/c=2016 cmrbc/a^2(3b+c)+ac/b^2(3c+a)+ab/c^2(3a+b)>=504

Answer 1

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\rightarrow\left(x;y;z\right)\) suy ra x, y, z >0 và x + y + z = 2016

BĐT \(\Leftrightarrow\frac{\frac{1}{yz}}{\frac{1}{x^2}\left(\frac{3}{y}+\frac{1}{z}\right)}+\frac{\frac{1}{zx}}{\frac{1}{y^2}\left(\frac{3}{z}+\frac{1}{x}\right)}+\frac{\frac{1}{xy}}{\frac{1}{z^2}\left(\frac{3}{x}+\frac{1}{y}\right)}\ge504\)

\(\Leftrightarrow\frac{x^2}{3z+y}+\frac{y^2}{3x+z}+\frac{z^2}{3y+x}\ge504\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel suy ra:

\(VT\ge\frac{\left(x+y+z\right)^2}{4\left(x+y+z\right)}=\frac{x+y+z}{4}=\frac{2016}{4}=504\) (đpcm)

Đẳng thức xảy ra khi x = y = z = 672 hay \(a=b=c=\frac{1}{672}\)