a,Cho a²+b²+c²+3=2(a+b+c )Chứng minh rằng a=b=c=1
b,Cho (a+b+c)²=3(ab+bc+ca)Chứng minh rằng a=b=c
c,Cho (a-b)²+(b-c)²+(c-a)²=(a+b-2c)²+(b+c-2a)²+(c+a-2b)²Chứng minh rằng a=b=c
a,Cho a²+b²+c²+3=2(a+b+c )Chứng minh rằng a=b=c=1
b,Cho (a+b+c)²=3(ab+bc+ca)Chứng minh rằng a=b=c
c,Cho (a-b)²+(b-c)²+(c-a)²=(a+b-2c)²+(b+c-2a)²+(c+a-2b)²Chứng minh rằng a=b=c
$\begin{array}{l} a)\\ {a^2} + {b^2} + {c^2} + 3 = 2\left( {a + b + c} \right)\\ \Leftrightarrow \left( {{a^2} – 2a + 1} \right) + \left( {{b^2} – 2b + 1} \right) + \left( {{c^2} – 2c + 1} \right) = 0\\ \Leftrightarrow {\left( {a – 1} \right)^2} + {\left( {b – 1} \right)^2} + {\left( {c – 1} \right)^2} = 0\\ \Leftrightarrow \left\{ \begin{array}{l} a – 1 = 0\\ b – 1 = 0\\ c – 1 = 0 \end{array} \right. \Leftrightarrow a = b = c = 1\\ b)\\ {\left( {a + b + c} \right)^2} = 3\left( {ab + bc + ca} \right)\\ \Leftrightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 3ab + 3bc + 3ca\\ \Leftrightarrow {a^2} – ab + {b^2} – bc + {c^2} – ca = 0\\ \Leftrightarrow 2{a^2} – 2ab + 2{b^2} – 2bc + 2{c^2} – 2ca = 0\\ \Leftrightarrow \left( {{a^2} – 2ab + {b^2}} \right) + \left( {{b^2} – 2bc + {c^2}} \right) + \left( {{c^2} – 2ca + {a^2}} \right) = 0\\ \Leftrightarrow {\left( {a – b} \right)^2} + {\left( {b – c} \right)^2} + {\left( {c – a} \right)^2} = 0\\ \Leftrightarrow \left\{ \begin{array}{l} a – b = 0\\ b – c = 0\\ c – a = 0 \end{array} \right. \Leftrightarrow a = b = c \end{array}$
c)
$\begin{array}{l} VT = {\left( {a – b} \right)^2} + {\left( {b – c} \right)^2} + {\left( {c – a} \right)^2}\\ VT = {\left( {a + b} \right)^2} – 4ab + {\left( {b + c} \right)^2} – 4bc + {\left( {c + a} \right)^2} – 4ac\left( 1 \right)\\ VP = {\left( {a + b – 2c} \right)^2} + {\left( {b + c – 2a} \right)^2} + {\left( {c + a – 2b} \right)^2}\\ = {\left[ {\left( {a + b} \right) – 2c} \right]^2} + {\left[ {\left( {b + c} \right) – 2a} \right]^2} + {\left[ {\left( {c + a} \right) – 2b} \right]^2}\\ = {\left( {a + b} \right)^2} – 4c\left( {a + b} \right) + 4{c^2} + {\left( {b + c} \right)^2} – 4a\left( {b + c} \right) + 4{a^2} + {\left( {c + a} \right)^2} – 4b\left( {c + a} \right) + 4{b^2}\left( 2 \right)\\ \left( 1 \right),\left( 2 \right) \Rightarrow – 4c\left( {a + b} \right) + 4{c^2} – 4a\left( {b + c} \right) + 4{a^2} – 4b\left( {c + a} \right) + 4{b^2} = – 4ab – 4bc – 4ca\\ \Rightarrow ab – \left( {a + b} \right)c + {c^2} + bc – \left( {b + c} \right)a + {a^2} + ac – \left( {a + c} \right)c + {b^2} = 0\\ \Rightarrow ab – ac – bc + {c^2} + bc – ab – ac + {a^2} + ac – ab – bc + {b^2} = 0\\ \Rightarrow {a^2} + {b^2} + {c^2} – ab – bc – ca = 0\\ \Rightarrow {\left( {a – b} \right)^2} + {\left( {b – c} \right)^2} + {\left( {c – a} \right)^2} = 0\\ \Rightarrow \left\{ \begin{array}{l} a – b = 0\\ b – c = 0\\ c – a = 0 \end{array} \right. \Leftrightarrow a = b = c \end{array}$
`a)`
` a^2 + b^2 + c^3 + 3 = 2(a+b+c)`
`\to a^2 + b^2 + c^3 + 3 -2a -2b -2c= 0`
`\to (a^2 -2a+1)+(b^2 -2b+1) + (c^2 -2c+1) = 0`
`\to (a-1)^2 + (b-1)^2 + (c-1)^2=0`
Ta có ` (a-1)^2 \ge0;\ (b-1)^2 \ge0;\ (c-1)^2 \ge 0`
` \to (a-1)^2 + (b-1)^2 + (c-1)^2 \ge 0\ ∀ a,b,c`
Dấu ` =` xảy ra khi ` a= b = c = 1`
`\to` Đpcm
`b)`
` (a+b+c)^2 = 3(ab+bc+ac)`
`\to a^2+b^2 +c^2 +2ab +2bc +2ac = 3ab + 3bc +3ac`
`\to a^2 +b^2 +c^2 – ab – bc – ac = 0`
`\to 2a^2 +2b^2 + 2c^2 – 2ab – 2bc – 2ac = 0`
`\to (a^2 -2ab + b^2)+(b^2 – 2bc + c^2)+(c^2 – 2ac +a^2) = 0`
`\to (a-b)^2 +(b-c)^2 + (a-c)^2 = 0`
`\to` $\begin{cases} a – b = 0 \\\\ b- c = 0 \\\\ c – a = 0 \end{cases}$
`\to a = b =c`
`\to` Đpcm
`c)`
` (a-b)^2 + (b-c)^2 + (c-a)^2 = (a+b-2c)^2 + (b+c -2a)^2 + (c+a-2b)^2`
`\to [ ( a-b)^2 – (a+b-2c)^2] + [ (b-c)^2 – (b+c -2a)^2] + [ (c-a)^2 – (c+a-2b)^2] = 0`
`\to (a-b+a+b-2c)(a-b-a-b+2c) + ( b-c+b+c-2a)(b-c-b-c+2a) + (c-a+c+a-2b)(c-a-c-a+2b) = 0`
`\to ( 2a -2c)(2c-2b) + ( 2b – 2a)(2a – 2c) + ( 2c – 2b)(2b – 2a) = 0`
`\to 2. [ (a-c)(c-b) + (b-a)(a-c) + (c-b)(b-a)] = 0`
`\to (a-c)(c-b) + (b-a)(a-c) + (c-b)(b-a) = 0`
`\to ac – ab – c^2 + bc + ab – bc – a^2 +ac + bc – ac – b^2 +ab = 0`
`\to (ac-ac+ac) + (ab-ab+ab) + (bc – bc+bc) – a^2 – b^2 – c^2 = 0`
`\to ab + ac + bc – a^2 – b^2 – c^2 = 0`
`\to a^2 + b^2 + c^2 – ab – ac – bc = 0`
`\to 2a^2 +2b^2 + 2c^2 – 2ab – 2bc – 2ac = 0`
`\to (a^2 -2ab + b^2)+(b^2 – 2bc + c^2)+(c^2 – 2ac +a^2) = 0`
`\to (a-b)^2 +(b-c)^2 + (a-c)^2 = 0`
`\to` $\begin{cases} a – b = 0 \\\\ b- c = 0 \\\\ c – a = 0 \end{cases}$
`\to a = b =c`
`\to` Đpcm