Question

cho a^2+b^2+c^2=ab+bc+ac

chứng minh rằng: a=b=c

Answer 1

Ta áp dụng Bđt Cô-si

\(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2\sqrt{a^2b^2}=2ab\left(1\right)\)

\(\left(b-c\right)^2\ge0\Leftrightarrow b^2+c^2\ge2\sqrt{b^2c^2}=2bc\left(2\right)\)

\(\left(a-c\right)^2\ge0\Leftrightarrow a^2+c^2\ge2\sqrt{a^2c^2}=2ac\left(3\right)\)

Cộng theo vế của (1),(2) và (3) có:

\(2\left(a^2+b^2+c^2\right)\ge2\left(ab+ac+bc\right)\)

\(\Rightarrow a^2+b^2+c^2\ge ab+ac+bc\)

Dấu = khi a=b=c

-->Đpcm

Answer 2

Ta có: \(a^2+b^2+c^2=ab+bc+ca\)

<=>\(a^2+b^2+c^2-ab-bc-ca=0\)

<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

<=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\left(a-b\right)^2\ge0,\left(b-c\right)^2\ge0,\left(c-a\right)^2\ge0\)

=>\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}< =>\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}}\)

Vậy a=b=c

Answer 3

Ta có : \(a^2+b^2+c^2=ab+bc+ca\)

\(\Rightarrow a^2+b^2-c^2-ab-bc-ca=0\)

\(\Rightarrow a^2-ab+b^2-bc+c^2-ca=0\)\(0\)

\(\Rightarrow a\left(a-b\right)+b\left(b-c\right)+c\left(c-a\right)=0\)

\(\Rightarrow a\left(a-b\right)=0\)             \(a=o\)

     \(b\left(b-c\right)=0\)\(\Rightarrow\)      \(b=0\)

      \(c\left(c-a\right)=0\)              \(c=0\)

\(\Rightarrow a=b=c\)

\(\Rightarrowđpcm\)

Answer 4

Ta có: a² + b² + c² = ab + bc + ca

=> 2(a² + b² + c²) = 2(ab + bc + ca)

=> 2a² + 2b² + 2c² = 2ab + 2bc + 2ca

=> a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ca = 0

=> (a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ca + a²) = 0

=> (a - b)² + (b - c)² + (c - a)² = 0

<=>(a - b)² = 0 ; (b - c)² = 0 ; (c - a)² = 0

=> a = b = c

Answer 5

Ta có: a^2 +b^2 +c^2=ab+bc+ac

suy ra 2*(a^2 +b^2 +c^2)=2*(ab+bc+ac)

suy ra 2a^2 +2b^2 +2c^2-2ab-2bc-2ac=0

suy ra (a-b)^2+(b-c)^2+(a-c)^2=0

suy ra a-b=0 , b-c=0 , a-c=0

suy ra a=b=c

Answer 6

Ta có:\(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c^2\right)+\left(c-a\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow}a=b=c}\)