Question

bài 1 cho a,b,c\(\ge1\)

chứng minh rằng

a(b+c)+b(c+a)+c(a+b)+2\(\left(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{1+c^2}\right)\ge9\)

bài 2 cho x,y,z>0 thỏa mãn \(x^3+y^3+z^3=3\)

Tìm min của A=3xy+3yz+3zx-xyz

Answer 1

bài 3

cho a,b,c>0. chứng minh rằng

\(\left(1+\dfrac{2a}{b}\right)^2+\left(1+\dfrac{2b}{c}\right)^2+\left(1+\dfrac{2c}{a}\right)^2\ge\dfrac{9\left(a+b+c\right)^2}{ab+bc+ca}\)

Answer 2

E.x 3:

Áp dụng bunyakovsky:

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

Áp dụng cauchy-schwarz:

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

Đặt \(\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}=t\) thì \(t\ge3\)

Cần chứng minh \(\dfrac{1}{3}\left(3+2t\right)^2\ge9t\Leftrightarrow\left(t-3\right)\left(4t-3\right)\ge0\)( đúng)

Vậy BĐT được chứng minh .