Câu hỏi của Ngo Hiệu

Question

Choa,b,c là ba số dương thỏa mãn:a+b+c=1.Chứng minh rằng:c+ab/a+b + a+bc/b+c + b+ac/a+c >=2

Nguyễn Thành Trương · Accepted Answer

Áp dụng BĐT AM-GM ta có:

$VT=\dfrac{c+ab}{a+b}+\dfrac{a+bc}{b+c}+\dfrac{b+ac}{a+c}$

$=\dfrac{c\left(a+b+c\right)+ab}{a+b}+\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ac}{a+c}$

$=\dfrac{ac+bc+c^2+ab}{a+b}+\dfrac{a^2+ab+ac+bc}{b+c}+\dfrac{ab+b^2+bc+ac}{a+c}$

$=\dfrac{\left(b+c\right)\left(c+a\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}$

$\ge2\left(a+b+c\right)=2\left(a+b+c=1\right)$

Khi $a=b=c=\dfrac{1}{3}$Câu trả lời: Áp dụng BĐT AM-GM ta có:

$VT=\dfrac{c+ab}{a+b}+\dfrac{a+bc}{b+c}+\dfrac{b+ac}{a+c}$

$=\dfrac{c\left(a+b+c\right)+ab}{a+b}+\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ac}{a+c}$

$=\dfrac{ac+bc+c^2+ab}{a+b}+\dfrac{a^2+ab+ac+bc}{b+c}+\dfrac{ab+b^2+bc+ac}{a+c}$

$=\dfrac{\left(b+c\right)\left(c+a\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}$

$\ge2\left(a+b+c\right)=2\left(a+b+c=1\right)$

Khi $a=b=c=\dfrac{1}{3}$

tran thi mai anh · Answer

BĐT AM-GM là gì vậyCâu trả lời: BĐT AM-GM là gì vậy