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\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
( a - b )2 + ( b - c )2 + ( c - a )2 = 4( a2 + b2 + c2 - ab - bc - ca )
<=> a2 - 2ab + b2 + b2 - 2bc + c2 + c2 - 2ca + a2 = 4( a2 + b2 + c2 - ab - bc - ca )
<=> 2( a2 + b2 + c2 - ab - bc - ca ) = 4( a2 + b2 + c2 - ab - bc - ca )
<=> 2( a2 + b2 + c2 - ab - bc - ca ) = 0 ( bớt 2( a2 + b2 + c2 - ab - bc - ca ) ở cả hai vế )
<=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0
<=> ( a2 - 2ab + b2 ) + ( b2 - 2bc + c2 ) + ( c2 - 2ca + a2 ) = 0
<=> ( a - b )2 + ( b - c )2 + ( c - a )2 = 0 (1)
Ta có : \(\hept{\begin{cases}\left(a-b\right)^2\\\left(b-c\right)^2\\\left(c-a\right)^2\end{cases}}\ge0\forall a,b,c\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Dấu "=" xảy ra ( tức (1) ) <=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\)
=> đpcm
Other way:\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(đúng)
Dấu "=" xảy ra khi a=b=c
a b<a+b> <a-b> + bc < b - c> < b + c >+ ca < c - a > < c + a>
a² b+ ab² + a² b - ab² + b² c -bc² +b² c + bc² + c² a -ca² + c² a +ca²
<a² b +a² b> + < ab² - ab² > + < b²c + b² c > + <-bc² + bc² > + < c² a +c² a> + <-ca² + ca² >
2 a² b + 2 b² c +2 c² a
XONG NHA NGƯỜI ANH EM
\(ab\left(a^2-b^2\right)+bc\left(b^2-c^2\right)+ca\left(c^2-a^2\right)\)
\(=a^3b-ab^3+b^3c-bc^3-ca\left(a^2-c^2\right)\)
\(=b\left(a^3-c^3\right)-b^3\left(a-c\right)-ca\left(a-c\right)\left(a+c\right)\)
\(=b\left(a-c\right)\left(a^2+ac+c^2\right)-b^3\left(a-c\right)-ca\left(a-c\right)\left(a+c\right)\)
\(=\left(a-c\right)\left(a^2b+abc+bc^2-b^3-a^2c-c^2a\right)\)
\(=\left(a-c\right)\left[b\left(a^2-b^2\right)+ac\left(b-a\right)+c^2\left(b-a\right)\right]\)
\(=\left(a-c\right)\left[b\left(a-b\right)\left(a+b\right)-ac\left(a-b\right)-c^2\left(a-b\right)\right]\)
\(=\left(a-c\right)\left(a-b\right)\left(ab+b^2-ac-c^2\right)\)
\(=\left(a-c\right)\left(a-b\right)\left[a\left(b-c\right)+\left(b-c\right)\left(b+c\right)\right]\)
\(=\left(a-c\right)\left(a-b\right)\left(b-c\right)\left(a+b+c\right)\)