Toán Lớp 8: Chứng minh các bđt: 3 (a ² + b ² + c ²) ≥ (a + b + c) ² ≥ 3(ab+bc+ca)

Question

Toán Lớp 8:  Chứng minh các bđt: 3 (a ² + b ² + c ²)  ≥ (a + b + c) ²  ≥  3(ab+bc+ca)

dinhcongson · Answer

Giải đáp:       Lời giải và giải thích chi tiết:   Ta c&oacute; (a+b+c)&sup2;=a&sup2;+b&sup2;+c&sup2;+2ab+2bc+2ac     3 (a &sup2; + b &sup2; + c &sup2;) = 3a&sup2;+3b&sup2;+3c&sup2;     &rArr;3 (a &sup2; + b &sup2; + c &sup2;) &ge; (a + b + c) &sup2;     &hArr;3a&sup2;+3b&sup2;+3c&sup2;&ge;a&sup2;+b&sup2;+c&sup2;+2ab+2bc+2ac     &hArr;2a&sup2;+2b&sup2;+2c&sup2;&ge;2ab+2bc+2ac     &hArr;2a&sup2;+2b&sup2;+2c&sup2;-2ab-2bc-2ac&ge;0     &hArr;(a&sup2;-2ab+b&sup2;)+(b&sup2;-2bc+c&sup2;)+(c&sup2;-2ac+a&sup2;) &ge;0     &hArr;(a-b)&sup2;+(b-c)&sup2;+(c-a)&sup2;&ge;0     M&agrave; (a-b)&sup2;&ge;0 ; (b-c)&sup2;&ge;0 ; (c-a)&sup2;&ge;0     &rArr;3 (a &sup2; + b &sup2; + c &sup2;) &ge; (a + b + c) &sup2; (1)     Lại  c&oacute; : (a + b + c) &sup2; &ge; 3(ab+bc+ca)     &hArr;a&sup2;+b&sup2;+c&sup2;+2ab+2bc+2ac &ge; 3ab+3bc + 3ac     &hArr;a&sup2;+b&sup2;+c&sup2;-ab-bc-ac&ge;0     &hArr;2.(a&sup2;+b&sup2;+c&sup2;-ab-bc-ac)&ge;0     &hArr;2a&sup2;+2b&sup2;+2c&sup2;&ge;2ab+2bc+2ac     &hArr;2a&sup2;+2b&sup2;+2c&sup2;-2ab-2bc-2ac&ge;0     &hArr;(a&sup2;-2ab+b&sup2;)+(b&sup2;-2bc+c&sup2;)+(c&sup2;-2ac+a&sup2;) &ge;0     &hArr;(a-b)&sup2;+(b-c)&sup2;+(c-a)&sup2;&ge;0     M&agrave; (a-b)&sup2;&ge;0 ; (b-c)&sup2;&ge;0 ; (c-a)&sup2;&ge;0     &rArr;(a + b + c) &sup2; &ge; 3(ab+bc+ca)(2)     Từ (1)(2)&rArr;đccm

myngocla · Answer

Giải đáp+Lời giải và giải thích chi tiết: 3(a^2+b^2+c^2)>=(a+b+c)^2>=3ab+3bc+3ca <=>3a^2+3b^2+3c^2>=a^2+b^2+c^2+2ab+2bc+2ca>=3ab+3bc+3c <=>{(3a^2+3b^2+3c^2>=a^2+b^2+c^2+2ab+2bc+2ca),(a^2+b^2+c^2+2ab+2bc+2ca>=3ab+3bc+3c):} <=>{(2a^2+2b^2+2c^2>=2ab+2bc+2ca),(a^2+b^2+c^2>=ab+bc+ca):} <=>2a^2+2b^2+2c^2>=2ab+2bc+2ca <=>a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2>=0 <=>(a-b)^2+(b-c)^2+(c-a)^2>=0 luôn đúng Dấu '=' xảy ra khi a=b=c