Question

[Lê Đức Anh]

cho \(0\le a,b,c\le2\)và a+b+c=3. CMR: a²+b²+c²\(\le\)5

Answer 1

[hung pham tien]

từ giả thuyết suy ra : abc >0

có 2>a,c,b ->> (2-a)(2-b)(2-c)\(\ge\)0

\(\Leftrightarrow\)8+2(ab+ac+bc) -4(a+b+c)-abc \(\ge\)0

\(\Leftrightarrow\)8+2(ab+ac+bc)-4.3-abc \(\ge\)0

\(\Leftrightarrow\)2(ab+ac+bc) \(\ge\)4+abc \(\ge\)4 (1)

Cộng a²+b²+c² vào (1)

2(ab+ac+bc)+a²+b²+c²\(\ge\)4+a²+b²+c²

(a+b+c)²-4\(\ge\)a²+b²+c²

^{thay a+b+c=3 vào}

9-4^\(\ge\)a²+b²+c²

^5 \(\ge\)a²+b²+c²

a²+b²+c² \(\le\)5

Answer 2

[nguyen duc vuong]

cauhc lop may