Question

Cho \(-1< a,b,c< 1;a+b+c=0\)

CMR:\(a^2+b^2+c^2< 2\)

Answer 1

Lời giải:

Vì $a,b,c< 1$ nên \((a-1)(b-1)(c-1)< 0\)

\(\Leftrightarrow abc-(ab+bc+ac)+(a+b+c)-1< 0\)

\(\Leftrightarrow ab+bc+ac>abc-1(1)\)

Lại có:

\(a,b,c>-1\Rightarrow (a+1)(b+1)(c+1)>0\)

\(\Leftrightarrow abc+ab+bc+ac+a+b+c+1>0\)

\(\Leftrightarrow ab+bc+ac> -abc-1(2)\)

Lấy \((1)+(2)\Rightarrow 2(ab+bc+ac)>-2\)

\(\Leftrightarrow a^2+b^2+c^2+2(ab+bc+ac)> -2+(a^2+b^2+c^2)\)

\(\Leftrightarrow (a+b+c)^2+2> a^2+b^2+c^2\)

\(\Leftrightarrow 2> a^2+b^2+c^2\) (đpcm)