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On the existence of inequalities For any real number x, the inequality │ x + 1 │ + X-2 │ > a holds, and the value range of real number a is obtained Analysis 2: using the absolute value inequality │ a │ - B │ ＜ a ± B │ ＜ a │ + B │ to solve the minimum value of F (x) = │ x + 1 │ + X-2 │ Let f (x) = x + 1 + X-2, x + 1 + X-2 = 3, f (x) min = 3. A < 3 Why can we do this?
As long as the equal sign can be obtained
Here is (x + 1) (X-2) ≥ 0
Obviously it can
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- 7. The position of the known real number AB on the number axis is shown in the figure: try to simplify: root (a-b) &# 178; - | a + B|
- 8. The positions of real numbers a and B on the number axis are shown in the figure. Try to simplify | a + B | + | A-B | + | ab| -----b--------0--------a--------》
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- 10. If two points a and B on the number axis represent real numbers a and B respectively, the length of line AB is () A. a-bB. a+bC. |a-b|D. |a+b|
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- 14. Prove the inequality: (A & sup2; + B & sup2;) (C & sup2; + D & sup2;) ≥ (AC + BD) & sup2;
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- 16. The proof of inequality (AC + BD) & sup2; ≤ (A & sup2; + B & sup2;) (C & sup2; + D & sup2;)
- 17. There are four inequalities: (1) A2 + B2 + C2 ≥ AB + BC + AC; (2) a (1-A) ≤ 14; (3) BA + ab ≥ 2; (4) (A2 + B2) (C2 + D2) ≥ (AC + BD) 2; where () A. 1 B. 2 C. 3 d. 4
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- 20. a. B, C, D are all real numbers. A group of values (a, B, C, d) which make the inequality AB > CD > 0 and ad < BC hold are______ (just write a set of values suitable for the condition)