Cho a,b,c>0 CMR
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}) \)
@Ace Legona giúp mình
Áp dụng BĐT Cauchy-Schwarz, ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{\left(1+1+1\right)^2}{a+b+b}=\dfrac{9}{a+2b}\)
\(\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{9}{b+2c}\)
\(\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{9}{c+2a}\)
Cộng vế theo vế rồi rút gọn, ta được
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\left(\text{đ}pcm\right)\)
Đẳng thức xảy ra khi \(a=b=c\)
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