Question

cho a,b,c>0 thỏa mãn : \(\frac{1}{x+1}+\frac{35}{35+2y}\le\frac{4z}{4z+57}\)

tìm min xyz

Answer 1

Áp dụng BĐT AM-GM ta có:

\(\frac{4z}{4z+57}\ge\frac{1}{1+x}+\frac{35}{35+2y}\ge2\sqrt{\frac{35}{\left(1+z\right)\left(35+2y\right)}}\)

\(\frac{x}{1+x}\ge\frac{57}{4z+57}+\frac{35}{35+2y}\ge2\sqrt{\frac{35\cdot57}{\left(4z+57\right)\left(35+2y\right)}}\)

\(\frac{2y}{35+2y}\ge\frac{57}{4z+57}+\frac{1}{1+x}\ge2\sqrt{\frac{57}{\left(4z+57\right)\left(1+x\right)}}\)

\(\Rightarrow8abc\ge8\cdot1995\Rightarrow abc\ge1995\)

Đẳng thức xảy ra khi \(x=2;y=35;z=\frac{57}{2}\)