Câu hỏi của Nguyet9ak47

Question

Cho a,b,c là độ dài 3 cạnh của một tam giác Chứng minh rằng :      nếu a+b+c=2 thì a2+b2+c2+2abc <2

Le Nhat Phuong · Accepted Answer

Nguyet9ak47 mk ko bt có đúng ko nhưng bn tham khảo nhé:ta co a+b>c suy ra 2cc<1,a<1,b<1 (1-a)(1-b)(1-c)>0 =>ab+bc+ac>1+abc lai co 4=2(ab+bc+ac)+a2+b2+c2 tu do suy ra 4>a2+b2+c2+2(1+abc)=>a2+b2+c2+2abc<2=>... a,b,c>0) P/s: Nguyet9ak47, Chứng minh rằng sao bn ko viết là CMRCâu trả lời: Nguyet9ak47 mk ko bt có đúng ko nhưng bn tham khảo nhé:ta co a+b>c suy ra 2cc<1,a<1,b<1 (1-a)(1-b)(1-c)>0 =>ab+bc+ac>1+abc lai co 4=2(ab+bc+ac)+a2+b2+c2 tu do suy ra 4>a2+b2+c2+2(1+abc)=>a2+b2+c2+2abc<2=>... a,b,c>0) P/s: Nguyet9ak47, Chứng minh rằng sao bn ko viết là CMR

©ⓢ丶κεη春╰‿╯ · Answer

Câu trả lời hay nhất: Tam giác ABC có độ dài 3 cạnh là a,b,c và có chu vi là 2 --> a + b + c = 2 Trong 1 tam giác thì ta có: a < b + c --> a + a < a + b + c --> 2a < 2 --> a < 1 Tương tự ta có : b < 1, c < 1 Suy ra: (1 - a)(1 - b)(1 - c) > 0 ⇔ (1 – b – a + ab)(1 – c) > 0 ⇔ 1 – c – b + bc – a + ac + ab – abc > 0 ⇔ 1 – (a + b + c) + ab + bc + ca > abc Nên abc < -1 + ab + bc + ca ⇔ 2abc < -2 + 2ab + 2bc + 2ca ⇔ a² + b² + c² + 2abc < a² + b² + c² – 2 + 2ab + 2bc + 2ca ⇔ a² + b² + c² + 2abc < (a + b + c)² - 2 ⇔ a² + b² + c² + 2abc < 2² - 2 , do a + b = c = 2 ⇔ a² + b² + c² + 2abc < 2 --> đpcm p/s: kham khảoCâu trả lời: Câu trả lời hay nhất: Tam giác ABC có độ dài 3 cạnh là a,b,c và có chu vi là 2 --> a + b + c = 2 Trong 1 tam giác thì ta có: a < b + c --> a + a < a + b + c --> 2a < 2 --> a < 1 Tương tự ta có : b < 1, c < 1 Suy ra: (1 - a)(1 - b)(1 - c) > 0 ⇔ (1 – b – a + ab)(1 – c) > 0 ⇔ 1 – c – b + bc – a + ac + ab – abc > 0 ⇔ 1 – (a + b + c) + ab + bc + ca > abc Nên abc < -1 + ab + bc + ca ⇔ 2abc < -2 + 2ab + 2bc + 2ca ⇔ a² + b² + c² + 2abc < a² + b² + c² – 2 + 2ab + 2bc + 2ca ⇔ a² + b² + c² + 2abc < (a + b + c)² - 2 ⇔ a² + b² + c² + 2abc < 2² - 2 , do a + b = c = 2 ⇔ a² + b² + c² + 2abc < 2 --> đpcm p/s: kham khảo

Lê Mạnh Hùng · Answer

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