Question

Cho a,b,c > 0 . CMR:

\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)

Answer 1

Đặt P=\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)

Không mất tính tổng quát giả sử a ≥b ≥ c , thế thì \(\dfrac{1}{b+c}\ge\dfrac{1}{c+a}\ge\dfrac{1}{a+b}\) .Áp dụng bất đẳng thức Chebyshev cho hai dãy đơn điệu cùng chiều ta có :

\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{1}{3}\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\)

\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\left(\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1\right)\)

Hay \(P\ge\dfrac{1}{3}\left(P+3\right)\) nghĩa là \(P\ge\dfrac{3}{2}^{\left(đpcm\right)}\)

Answer 2

https://i.imgur.com/UCvj2rx.jpg

Answer 3

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