<=>2ab+2bc+2ca<=1=1^2=(a+b+c)^2

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

<=>a^2+b^2+c^2>=0

a,b,c khong dong thoi =0

=> dang thuc khong xay ra

=> ab+bc+ca<1/2=>dpcm

(a+b+c)=1

a^2+b^2+c^2+2ab+2bc+2ca=1

a^^2+b^2+c^2>=0

=>2ab+2bc+2ca<=1

Đẳng thức khi (a+b+c=1 &0=> vô nghiệm

=> 2ab+2bc+2ca<1

=>ab+2bc+2ca<1/2

=>đpcm

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)

Ta có: \(a^2+b^2+c^2=ab+bc+ca\)

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

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

\(\Leftrightarrow a=b=c\)

Ta có:

\(a+b+c=1\)

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

\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=1\)

\(\Rightarrow2ab+2ac+2bc=1-a^2-b^2-c^2\)

\(\Rightarrow2\left(ab+ac+bc\right)=1-a^2-b^2-c^2\)

Vì \(1-a^2-b^2-c^2< 1\)

\(\Rightarrow2\left(ab+ac+bc\right)< 1\)

\(\Rightarrow ab+ac+bc< \dfrac{1}{2}\)

a + b + c =1 ⇔ (a + b + c)² = 1

⇔ a² + b² + c² + 2ab +2ac +2bc = 1

⇔2(ab + bc +ca) = 1 - a² + b² + c²

⇒2(ab + bc + ca) < 1

⇔ ab + bc +ca < \(\dfrac{1}{2}\)

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

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

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

ta có : \(a^2+b^2+c^2=ab+bc+ca\)

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

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

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

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

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

thấy có chỗ chưa hợp lý lắm ă :>

non vãi loonf đến câu này còn đéo bt ko bt đi học để làm gì

đúng trẻ trâu

\(a+b+c+ab+bc+ca=6abc\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z+xy+yz+zx=6\)

Ta cần chứng minh: \(x^2+y^2+z^2\ge3\)

Thật vậy:

\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)

\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\)

Cộng vế với vế:

\(3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge12\)

\(\Rightarrow x^2+y^2+z^2\ge3\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;1\right)\) hay \(\left(a;b;c\right)=\left(1;1;1\right)\)