\(a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ca\) 

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

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

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

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

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

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

Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(a-c\right)^2\ge0\\\left(b-c\right)^2\ge0\end{cases}}\Rightarrow\)\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\)

Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)

Vậy a = b = c (đpcm)

a² + b² + c² = ab + bc + ca

<=> 2(a² + b² + c²) = 2(ab + bc + ca)

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

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

<=> a = b = c

Ta có: a² + b² + c² - ab - bc - ca = 0

=> aa + bb + cc - ab - bc - ca = 0

=> aa + ab - bb + bc - cc -+ca = 0

=> a - b - c      = 0

=> a = b = c (đpcm)

a²+b²+c²-ab-bc-ca=2a²+2b²+2c²-2ab-2bc-2ca=a²-2ab+b²+b²-2bc+c²+c²-2cb+b²=(a-b)²+(b-c)²+(c-a)²=0

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

ta có \(a^2+b^2+c^2-ab+bc-ba=0\)

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

Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\)\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow a=b=c}\)

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

\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Rightarrow2a^2+2b^2+2c^2-\left(2ab+2bc+2ca\right)=0\)

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

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)    (1)

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

Nên (1) \(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)     \(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)

\(\Leftrightarrow a=b=c\left(đpcm\right)\)

Bình phương 2 vế ta được

2a²+2b²+2c²=2ab+2bc+2ac

Lấy VT trừ VP ta được

(a-b)²+(b-c)²+(c-a)²=0

=>a=b=c=0

a² + b² + c² = ab + bc + ca

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

<=> 2a² + 2b² + 2c² = 2ab + 2bc + 2ca

<=> 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)

\(\hept{\begin{cases}\left(a-b\right)^2\\\left(b-c\right)^2\\\left(c-a\right)^2\end{cases}\ge}0\forall a,b,c\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)

Dấu "=" xảy ra ( tức (1) ) <=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\left(đpcm\right)\)

b) Ta có : a\(^2\)+ b\(^2\)+ c\(^2\) =ab+bc+ca

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

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

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

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

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

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

<=> hoặc a-b=0 hoặc b-c=o hoặc a-c=o <=>a=b hoặc b=c hoặc a=c

=>a=b=c (đpcm)

a) Theo đề bài: \(a^2+b^2=ab\)

=>\(a^2+b^2-ab=0\)

=>\(a^2-2ab+b^2+ab=0\)

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

Vì \(\left(a-b\right)^2\ge0\)  để \(\left(a-b\right)^2+ab=0\) <=> \(\left(a-b\right)^2=ab=0\)

(a-b)²=0 <=> a-b=0 <=> a=b (đpcm)

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

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

=>\(2a^2+2b^2+2c^2=2ab+2bc+2ac\)

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

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

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

Vì \(\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(a-c\right)^2\ge0\end{cases}\) để \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

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

<=>a-b=b-c=a-c=0

<=>a=b=c (đpcm)

Áp dụng bất đẳng thức tam giác có a+b>c

                                                            <=>ac+bc > c^{2  (c>0)}

<=>a+b
   Tương tự có:ab+cb>b²    ac+ab >a²ab+bc>b2,ac+ab>a2

Cộng các bất đẳng thức trên ra điều phải chứng minh

2(a²+b²+c²)-2(ab+bc+ac)<a²+b²+c²<=>2(a²+b²+c²)>a²+b²+c² (dpcm)

ta có (ab+ac)/2 = (ba+bc)/3 = (ca+cb)/4 

=ab+ac-ba-bc+ca+cb/2-3+4 = 2ac/3

=ab+ac+ba+bc-ca-cb/2+3-4 = 2ab

=ab+ac-ba-bc-ca-cb/2-3-4 = 2bc/5

=> 2ac/3=2ab=2bc/5

Ta có 2ac/3=2ab/1 =>c/3 = b/1 => c/15 = b/5    (1)

          2ac/3 = 2bc/5 => a/3 = b/5                         (2)

từ (1) và(2) => a/3 = b/5 = c/15

bạn 2-3-4=5 ??