đặt b+c-a=x
a+c-b=y
a+b-c=z
ta có x+y=2c
x+z=2b
z+y=2a
ta lại có
2A=\(\dfrac{2a}{x}+\dfrac{2b}{y}+\dfrac{2c}{z}\)
2A=\(\dfrac{z+y}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)
2A=\(\dfrac{z}{x}+\dfrac{y}{x}+\dfrac{x}{y}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{y}{z}\)
2A=\(\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge2+2+2=6\)
=>2A= \(\dfrac{2a}{x}+\dfrac{2b}{y}+\dfrac{2c}{z}\ge6\)
<=>A≥3 (chia cả 2 vế cho 2 ) (đpcm)
Xin góp thêm cách nữa:
Am-Gm thẳng cho 3 số:
\(\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}\ge3\sqrt[3]{\dfrac{abc}{\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)}}\)
việc còn lại chỉ việc chứng minh :
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
Áp dụng BĐT Am-Gm ta có:
\(\left(a+b-c\right)\left(b+c-a\right)\le\dfrac{1}{4}\left(a+b-c+b+c-a\right)=b^2\)
\(\left(b+c-a\right)\left(c+a-b\right)\le c^2\)
\(\left(c+a-b\right)\left(a+b-c\right)\le a^2\)
Nhân lại ta có đpcm.Dấu = xảy ra khi a=b=c