Question

cho \(a,b,c>0\),\(abc=1\)

chứng minh rằng:

\(\sum\dfrac{1}{a^4\left(b+c\right)^2}\ge\dfrac{3}{4}\)

Answer 1

\(\Sigma\dfrac{1}{a^4\left(b+c\right)^2}=\Sigma\dfrac{a^2b^2c^2}{a^4\left(b+c\right)^2}=\Sigma\dfrac{\left(bc\right)^2}{\left(ab+ac\right)^2}\)

Đặt : \(ab=x;bc=y;ac=z\)

\(\Rightarrow\Sigma\dfrac{\left(bc\right)^2}{\left(ac+ab\right)^2}=\Sigma\dfrac{x^2}{\left(y+z\right)^2}=\Sigma\left(\dfrac{x}{y+z}\right)^2\)

Đặt \(\dfrac{x}{y+z}=n\); \(\dfrac{y}{z+x}=n\); \(\dfrac{z}{x+y}=k\)

\(\Rightarrow\Sigma\left(\dfrac{x}{y+z}\right)^2=m^2+n^2+k^2\)

Theo BĐT Nezbit

\(\Rightarrow n+m+k\ge\dfrac{3}{2}\)

Áp dụng BĐT Bunhiacopxki: \(m^2+n^2+k^2\ge\dfrac{\left(m+n+k\right)^2}{3}\ge\dfrac{\left(\dfrac{3}{2}\right)^2}{3}=\dfrac{3}{4}\)
=> ĐPCM

Answer 2

@Ace Legona @Ace legona