Question

1, a^2+b^2+c^2 >= ab + bc + ca 2, ( a+b+c)*(1/a + 1/b + 1/c) >= 9 3, a/b +b/c + c/a >= 0 a,b,c>0

Answer 1

\(1,\text{Giả sử }a^2+b^2+c^2\ge ab+bc+ca\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(\text{luôn đúng}\right)\)

Vậy \(a^2+b^2+c^2\ge ab+bc+ca\)

Dấu \("="\Leftrightarrow a=b=c\)

\(2,\forall a,b,c>0\\ \text{Áp dụng BĐT cosi: }\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\dfrac{1}{abc}}=9\sqrt[3]{\dfrac{abc}{abc}}=9\)

Dấu \("="\Leftrightarrow a=b=c\)