áp dụng BĐT cô si cho 2 số ta có

\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{bc}{a}.\dfrac{ac}{b}}\)

⇔\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{c^2}=2c\)

TT ta có \(\dfrac{ac}{b}+\dfrac{ab}{c}\ge2a\)

\(\dfrac{ab}{c}+\dfrac{bc}{a}\ge2b\)

cộng từng vế 3 BĐT trên

\(2\left(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\right)\ge2\left(a+b+c\right)\)

⇔ \(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\) (đpcm)

bạn nói hay;

cho tui tick.

140

tick nha mấy bn thank you very much

1234

ĐKXĐ x ≠1

\(B=\dfrac{3x^2-8x+6}{x^2-2x+1}\)

= \(\dfrac{\left(2x^2-4x+2\right)+\left(x^2-4x+4\right)}{x^2-2x+1}\)

= \(\dfrac{2\left(x^2-2x+1\right)}{x^2-2x+1}+\dfrac{x^2-4x+4}{x^2-2x+1}\)

= \(2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\)

do (x-2)² ≥0 ∀x

(x-1)² ≥0 ∀x

=> \(\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge0\)

<=> \(2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)

<=> B ≥ 2

Min B =2 khi

(x-2)² =0

⇔x-2=0

⇔x=2

vậy GTNN B =2 khi x=2