$\\$
`B = 3 |x-1| + |3x+5| +2014`
`-> B = |3| |x-1| + |3x+5| + 2014`
`-> B = |3 (x-1) | + |3x+5| + 2014`
`-> B = |3x-3| + |3x+5|+2014`
`-> B = |3x+5| + |3-3x| + 2014`
Áp dụng BĐT `|a| + |b| ≥ |a+b|` có :
`-> |3x+5| +|3-3x| ≥ |3x+5 + 3-3x| ≥ |8|=8∀x`
`-> |3x+5| + |3-3x| + 2014 ≥ 2014 + 8=2022∀x`
`-> B ≥2022∀x`
Dấu "`=`" xảy ra khi :
`↔ (3x+5) (3-3x) ≥0`
`↔` \(\left[ \begin{array}{l}\left\{ \begin{array}{l}3x+5≥0 \\3-3x ≥ 0\end{array} \right.\\ \left\{ \begin{array}{l}3x+5≤0 \\3-3x≤ 0 \end{array} \right.\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x≥\dfrac{-5}{3} \\x≤ 1 \end{array} \right. \text{(Luôn đúng)}\\ \left\{ \begin{array}{l}x≤\dfrac{-5}{3} \\x ≥1\end{array} \right. \text{(Loại)}\end{array} \right.\)
`↔ (-5)/3 ≤x≤1`
Vậy `min B=2022 ↔(-5)/3 ≤x≤1`
