Question

[Đạt Trần Tiến]

Cho 3 số dương a,b,c tm: a+b+c+ab+ca+bc=6abc

CMR: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{3}\)

@Lightning Farron

Answer 1

[Lightning Farron]

Sửa \(\dfrac{1}{3}\rightarrow3\)

Từ \(a+b+c+ab+bc+ca=6abc\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=6\)

Ta có: \(\dfrac{1}{a^2}+1\ge\dfrac{2}{a};\dfrac{1}{b^2}+1\ge\dfrac{2}{b};\dfrac{1}{c^2}+1\ge\dfrac{2}{c}\)

Và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge\dfrac{2}{ab};\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{2}{bc};\dfrac{1}{c^2}+\dfrac{1}{a^2}\ge\dfrac{2}{ac}\)

Cộng theo vế các BĐT trên ta có:

\(3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

\(\Leftrightarrow3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge12\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\ge4\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\)

\("="\Leftrightarrow a=b=c=1\)

Answer 2

[Nguyễn Huy Thắng]

để sau nha giờ bận .-.