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Làm đơn giản thế này thôi nhé An Kì :
Ta có : \(2016a+bc=\left(a+b+c\right)a+bc=a^2+ab+ac+bc=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)Tương tự : \(2016b+ac=\left(a+b\right)\left(b+c\right)\)
\(2016c+ab=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow\left(2016a+bc\right)\left(2016b+ac\right)\left(2016c+ab\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)
\(\frac{a}{ab+a+2016}+\frac{b}{bc+b+1}+\frac{2016c}{ac+2016c+2016}\)
\(=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{abc^2}{ac+abc^2+abc}\)
\(=\frac{a}{a.\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{abc^2}{ac.\left(1+bc+b\right)}\)
\(=\frac{1}{b+bc+1}+\frac{b}{b+bc+1}+\frac{bc}{b+bc+1}\)
\(=\frac{1+b+bc}{b+bc+1}=1\)
a)a2+b2+c2+3=2(a+b+c)
=>a2+b2+c2+1+1+1-2a-2b-2c=0
=>(a2-2a+1)+(b2-2b+1)+(c2-2c+1)=0
=>(a-1)2+(b-1)2+(c-1)2=0
=>a-1=b-1=c-1=0 <=>a=b=c=1
-->Đpcm
b)(a+b+c)2=3(ab+ac+bc)
=>a2+b2+c2+2ab+2ac+2bc -3ab-3ac-3bc=0
=>a2+b2+c2-ab-ac-bc=0
=>2a2+2b2+2c2-2ab-2ac-2bc=0
=>(a2- 2ab+b2)+(b2-2bc+c2) + (c2-2ca+a2) = 0
=>(a-b)2+(b-c)2+(c-a)2=0
Hay (a-b)2=0 hoặc (b-c)2=0 hoặc (a-c)2=0
=>a-b hoặc b=c hoặc a=c
=>a=b=c
-->Đpcm
c)a2+b2+c2=ab+bc+ca
=>2(a2+b2+c2)=2(ab+bc+ca)
=>2a2+2b2+c2=2ab+2bc+2ca
=>2a2+2b2+c2-2ab-2bc-2ca=0
=>a2+a2+b2+b2+c2+c2-2ab-2bc-2ca=0
=>(a2-2ab+b2)+(b2-2bc+c2)+(a2-2ca+c2)=0
=>(a-b)2+(b-c)2+(a-c)2=0
Hay (a-b)2=0 hoặc (b-c)2=0 hoặc (a-c)2=0
=>a-b hoặc b=c hoặc a=c
=>a=b=c
-->Đpcm
Giải :
a2 + b2 + c2 = ab + ac + bc
\(\Rightarrow\)2a2 + 2b2 + 2c2 = 2ab + 2ac + 2bc
\(\Rightarrow\)2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc = 0
\(\Rightarrow\)( a2 - 2ab + b2 ) + ( a2 - 2ac + c2 ) + ( b2 - 2bc + c2 ) = 0
\(\Rightarrow\)( a - b )2 + ( a - c )2 + ( b - c )2 = 0
Vì ( a - b )2 \(\ge\)0 với mọi a , b ; ( a - c )2 \(\ge\)với mọi a , c ; ( b - c )2 \(\ge\)0 với mọi b , c
Do đó ( a - b )2 + ( a - c )2 + ( b - c )2 = 0 khi a - b = a - c = b - c = 0
\(\Rightarrow\)a = b = c
ta có \(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\)
tương tự ta có
\(b^2+c^2\ge2bc;c^2+a^2\ge2ac\)
cộng từng vế của 3 bđt cùng chiều ta có
\(a^2+b^2+c^2\ge ab+bc+ca\)
dấu = xảy ra <=> a=b=c(ĐPCM)
Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=2\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
\(\Leftrightarrow-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2=0\)
Vì \(\left\{{}\begin{matrix} -\left(a-b\right)^2\le0\\-\left(b-c\right)^2\le0\\-\left(c-a\right)^2\le0\end{matrix}\right.\Rightarrow-\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\le0\)
Dấu ''= '' xảy ra \(\Leftrightarrow a=b=c\)
Vậy với a=b=c thì \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Ta có: \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-3ab-3bc-3ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ac-2bc-2ab=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow a=b=c\)
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
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
Ta có : \(2016a+bc=\left(a+b+c\right).a+bc=a^2+ab+ac+bc=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
\(2016b+ac=\left(a+b+c\right).b+ac=ab+b^2+bc+ac=b\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(b+c\right)\)
\(2016c+ab=\left(a+b+c\right)c+ab=ac+bc+c^2+ab=a\left(b+c\right)+c\left(b+c\right)=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow\left(2016a+bc\right)\left(2016b+ac\right)\left(2016c+ab\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\) (đpcm)