Ta có:
`|x-1|+|x+4|+|x-9|+|x-3|`
`=|-(x-1)|+|x+4|+|-(x-9)|+|x-3|`
`=|1-x|+|x+4|+|9-x|+|x-3|`
`=(|1-x|+|x-3|)+(|x+4|+|9-x|)`
Áp dụng BĐT: `|a|+|b|>=|a+b|` ta được:
`|1-x|+|x-3|>=|1-x+x-3|=|-2|=2`
`|x+4|+|9-x|>=|x+4+9-x|=|13|=13`
Do đó: `|1-x|+|x-3|+|x+4|+|9-x|>=2+13=15`
Hay: `|x-1|+|x+4|+|x-9|+|x-3|>=15` (đpcm)
∣x−1∣+∣x+4∣+∣x−9∣+∣x−3∣
\(= \mid - \left(\right. x - 1 \left.\right) \mid + \mid x + 4 \mid + \mid - \left(\right. x - 9 \left.\right) \mid + \mid x - 3 \mid\)
\(= \mid 1 - x \mid + \mid x + 4 \mid + \mid 9 - x \mid + \mid x - 3 \mid\)
\(= \left(\right. \mid 1 - x \mid + \mid x - 3 \mid \left.\right) + \left(\right. \mid x + 4 \mid + \mid 9 - x \mid \left.\right)\)
Áp dụng BĐT: \(\mid a \mid + \mid b \mid > = \mid a + b \mid\) ta được:
\(\mid 1 - x \mid + \mid x - 3 \mid > = \mid 1 - x + x - 3 \mid = \mid - 2 \mid = 2\)
\(\mid x + 4 \mid + \mid 9 - x \mid > = \mid x + 4 + 9 - x \mid = \mid 13 \mid = 13\)
Do đó: \(\mid 1 - x \mid + \mid x - 3 \mid + \mid x + 4 \mid + \mid 9 - x \mid > = 2 + 13 = 15\)
Hay: \(\mid x - 1 \mid + \mid x + 4 \mid + \mid x - 9 \mid + \mid x - 3 \mid > = 15\) (đpcm)
