Prove the inequality: (A & sup2; + B & sup2;) (C & sup2; + D & sup2;) ≥ (AC + BD) & sup2;
It is proved that: (A & sup2; + B & sup2;) (C & sup2; + D & sup2;) = A & sup2; C & sup2; + A & sup2; D & sup2; + B & sup2; D & sup2; ∵ A & sup2; D & sup2; + B & sup2; C & sup2; = (AD + BC) & sup2; - 2abcd ≥ 0 (AD + BC) & sup2; ≥ 2abcda & sup2; D & sup2; + B & sup2
RELATED INFORMATIONS
- 1. It is proved that for any ABCD, there is always (AC + BD) & sup2; ≤ (A & sup2; + B & sup2;) (C & sup2; + D & sup2;)
- 2. It is proved that for any a, B, C, D ∈ R, there is an inequality (AC + BD) ≤ (a + b) (c + D),
- 3. Given C > 0, let P: function y = C ^ X be a decreasing function on R; Q: inequality x + | x-2c | >
- 4. 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?
- 5. What's the difference between the solution of inequality and the existence of inequality? For example, is x greater than 1-B / a the same as x greater than 1-B / a? Good and quick answer,
- 6. In △ ABC, point D is the intersection of bisector of ∠ C and base ab. the following inequality is proved: CD & # 178; = AC · BC-AD · BD
- 7. Three inequalities are known: ① AB > 0. ② BC ad > 0. ③ C / a > D / b. taking two of them as conditions and the remaining one as conclusion, the number of correct propositions that can be formed is
- 8. If a > b, C
- 9. If A-B = 1 + √ 3, B-C = 1 - √ 3, find the value of 1 / (A & # 178; + B & # 178; + C & # 178; - AB AC BC)
- 10. If point C is the golden section of line AB, and AC
- 11. It is proved that for any a, B, C and D, they belong to R, and there is an inequality (AC + BD) ^ 2
- 12. The proof of inequality (AC + BD) & sup2; ≤ (A & sup2; + B & sup2;) (C & sup2; + D & sup2;)
- 13. 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
- 14. Three inequalities are known: (1) AB > 0, (2) C / a > D / B, (3) BC > ad If two of them are taken as the conditions and the other one is taken as the conclusion, the composition can be formed______ A correct proposition
- 15. Three inequalities are known: (1) AB > 0; (2) C / a > - D / B; (3) BC > ad If two of them are taken as conditions and the remaining one as a conclusion, a correct proposition can be formed Wrong number~ Three inequalities are known: (1) AB > 0; (2) C / AAD
- 16. 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)
- 17. If a, B, C and D are all real numbers, a set of inequalities (a, B, C, d) which make a / b > C / d > 0 and ad < BC hold are
- 18. If a, B, C and D are all real numbers, we try to list a set of values that make inequality a / b > C / d > 0 and ad > BC hold
- 19. Let a ^ n * (a ^ 2-B * c) + B ^ n (b ^ 2-ac) + C ^ n (C ^ 2-AB) > = 0 Prove a ^ n * (a ^ 2-B * c) + B ^ n (b ^ 2-ac) + C ^ n (C ^ 2-AB) > = 0, where n is any positive number
- 20. A, B and C are positive real numbers. Prove BC / A + AC / B + AB / C > = a + B + C