Question

cho a,b,c>0, chứng minh:

1)ab+bc+ca >= a√ab+b√ca+c√ab

2)a^2+b^2+c^2 >= a√ab+b√ca+c√ab

Answer 1

1, Áp dụng BĐT cosi cho a,b,c>0

\(ab+bc\ge2\sqrt{ab^2c}=2b\sqrt{ac}\\ bc+ca\ge2\sqrt{abc^2}=2c\sqrt{ab}\\ ca+ab\ge2\sqrt{a^2bc}=2a\sqrt{bc}\)

Cộng VTV 3 BĐT trên:

\(\Leftrightarrow2\left(ab+bc+ac\right)\ge2\left(b\sqrt{ac}+a\sqrt{bc}+c\sqrt{ab}\right)\\ \Leftrightarrow ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)

Answer 2

\(2,\)

Ta có

 \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\\ \Leftrightarrow a^2+b^2+c^2-ab-ac-bc\ge0\\ \Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

Áp dụng BĐT cm ở câu 1

Suy ra đpcm