a2 + b2 + c2 + 3 = 2(a + b + c)
=> a2 + b2 + c2 + 3 = 2a + 2b + 2c
=> a2 - 2a + 1 + b2 - 2b + 1 + c2 - 2c + 1 = 0
=> (a - 1)2 + (b - 1)2 + (c - 1)2 = 0
=> a - 1 = 0; b - 1 = 0; c - 1 = 0
=> a = 1; b = 1; c = 1 (đpcm)
Ta có : \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Rightarrow\) \(a^2+b^2+c^2-2a-2b-2c=0\)
\(\Rightarrow\) \(a^2-2a+1+b^2-2b+1+c^2-2c+1=0\)
\(\Rightarrow\) \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Do Vế Trái không âm
\(\Rightarrow\) \(\hept{\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}}\) \(\Rightarrow\) \(\hept{\begin{cases}a=1\\b=1\\c=1\end{cases}}\)
\(\Rightarrow\) \(a=b=c=1\)
\(\Rightarrow\) \(đpcm\)
Ta có \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
=> \(a^2-2a+1+b^2-2b+1+c^2-2c+1=0\)
=> \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
=> \(\left(a-1\right)^2=\left(b-1\right)^2=\left(c-1\right)^2=0\)
=> \(a-1=b-1=c-1=0\)
=> \(a=b=c=1\)
ta có: a2 + b2 + c2 + 3 = 2.(a+b+c)
=> a2 + b2 + c2 + 3 - 2.(a+b+c) = 0
a2 + b2 + c2 + 3 - 2a - 2b - 2c = 0
a - 2.1.a + 1 + b2 - 2.1.b +1 + c2 - 2.1.c = 0 ( tách 3 = 1 + 1 +1)
(a-1)2 + (b-1)2 + (c-1)2 = 0
mà \(\left(a-1\right)^2\ge0;\left(b-1\right)^2\ge0;\left(c-1\right)^2\ge0.\)
Dấu "=" xảy ra khi
(a-1)2 = 0 => a - 1= 0 => a = 1
(b-1)^2=0=> b - 1 = 0 => b = 1
(c-1)^2 = 0 => c -1 = 0 => c = 1
=> a = b =c = 1
\(a^2+b^2+c^2+3=2.\left(a+b+c\right)\)
\(a^2+b^2+c^2-2a-2b-2c+1+1+1=0\)
\(\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)^2=0\)
\(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Ta có: \(\hept{\begin{cases}\left(a-1\right)^2\ge0\forall a\\\left(b-1\right)^2\ge0\forall b\\\left(c-1\right)^2\ge0\forall c\end{cases}\Rightarrow}\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\forall a;b;c\)
Mà \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=1\\c=1\end{cases}\Rightarrow}}a=b=c=1\)
đpcm
Tham khảo nhé~
