Question

Cho ba số a, b, c >0,

Chứng minh rằng: a/b² +b/c²⁺c/a² ≥ 1/a+1/b+1/c

Answer 1

Ta có:

\(\dfrac{a}{b^2}+\dfrac{1}{a}\ge2\sqrt{\dfrac{a}{ab^2}}=\dfrac{2}{b}\)

\(\dfrac{b}{c^2}+\dfrac{1}{b}\ge2\sqrt{\dfrac{b}{bc^2}}=\dfrac{2}{c}\)

\(\dfrac{c}{a^2}+\dfrac{1}{c}\ge2\sqrt{\dfrac{c}{ca^2}}=\dfrac{2}{a}\)

Cộng vế:

\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\)

\(\Rightarrow\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)