2x+2|x|≥2√2 . . . (i) Case I: x≥0, then Eq. (i) becomes ‌2x+2x≥2√2 ‌⇒2x≥√2⇒x≥‌
1
2
Case II: x<0, then Eq. (i) becomes ‌2x+2−x≥2√2 ‌⇒t+‌
1
t
≥2√2,‌ where ‌2x=t ‌⇒t2−2√2t+1≥0 ‌⇒x≤log2(√2−1) Also, 0<√2−1<1,log2(√2−1)<0. ∴ The solution is (−∞,log2(√2−1))]∪[‌