Question

cho a,b,c,d > 0

cm: \(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge\frac{64}{a+b+c+d}\)

Answer 1

Áp dụng BĐT Cauchy -Schwarz dạng cộng mẫu thôi:

\(\text{VT}=\frac{1^2}{a}+\frac{1^2}{b}+\frac{2^2}{c}+\frac{4^2}{d}\geq \frac{(1+1+2+4)^2}{a+b+c+d}=\frac{64}{a+b+c+d}=\text{VP}\)

Dấu bằng xảy ra khi \(a=b=\frac{c}{2}=\frac{d}{4}>0\)

Answer 2

áp dụng BĐT cauchy-schwazs:

\(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge\frac{\left(1+1+2+4\right)^2}{a+b+c+d}=\frac{64}{a+b+c+d}\)

dấu = xảy ra khi \(\frac{1}{a}=\frac{1}{b}=\frac{2}{c}=\frac{4}{d}\Leftrightarrow a=b=\frac{c}{2}=\frac{d}{4}\)