Prove or disprove : for every real number x, |x − 8| + x ≥ 6.

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If $x>8$ , then $|x-8|+x=x-8+x=2x-8>2\cdot 8-8=8$ .

If $x\leq 8$ , then $|x-8|+x=8-x+x=8$ .

We have that $| x-8|+x\geq 8$ .

Since $8>6$ , it follows that $|x-8|+x\geq 6$ for every real number $x$ .

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