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Áp dụng BĐT Cauchy-Schwarz ta có:
\(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\Rightarrow\left(a+b+c\right)^2\le9\Rightarrow a+b+c\le3\left(1\right)\)
Ta có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow ab+bc+ca\le3\left(2\right)\)
Cộng vế với vế của\(\left(1\right),\left(2\right)\)ta được:
\(a+b+c+ab+bc+ca\le3+3=6\left(đpcm\right)\)
a, \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
=> a=b=c
\(a^2+b^2>=2ab\)
\(b^2+c^2>=2bc\)
\(a^2+c^2>=2ac\)
=> \(2\left(a^2+b^2+c^2\right)>=2\left(ab+bc+ac\right)\)DẤU '=' xảy ra khi a=b=c
\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)(1)
Vì \(\left(a-b\right)^2\ge0\); \(\left(b-c\right)^2\ge0\); \(\left(c-a\right)^2\ge0\)với \(\forall a,b,c\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(2)
Từ (1) và (2) \(\Rightarrow\)Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\)( đpcm )
Có ab + bc + ca = 0
=> 2ab + 2bc + 2ca = 0
Lại có a2 + b2 + c2 = 0 (1)
=> a2 + 2ab + b2 + 2bc + c2 + 2ca = 0
=> (a + b + c)2 = 0
=> a + b + c = 0 (2)
Từ (1) và (2) => a = b = c (đpcm)
Ta có: \(\hept{\begin{cases}a^2+b^2+c^2=0\\ab+bc+ca=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2a^2+2b^2+2c^2=0\\2ab+2bc+2ca=0\end{cases}}\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0;\forall a,b,c\\\left(b-c\right)^2\ge0;\forall a,b,c\\\left(c-a\right)^2\ge0;\forall a,b,c\end{cases}}\)\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0;\forall a,b,c\)
Do đó \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)
\(\Leftrightarrow a=b=c\left(đpcm\right)\)
Đề có sai ko bạn ?
Ta có: \(0\le\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)(1)
theo đề bài:
\(a^2+b^2+ab+bc+ac< 0\)
=> \(2\left(a^2+b^2+ab+bc+ac\right)< 0\)
=> \(2a^2+2b^2+2ab+2bc+2ac< 0\)(2)
Từ (1); (2) =>\(2a^2+2b^2+2ab+2bc+2ac< \) \(a^2+b^2+c^2+2ab+2bc+2ac\)
=> \(a^2+b^2< c^2\)