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Question

Với a, b,c là các số dương thoả mãn điều kiện a + b + c + ab + bc + ac = 6abc. Chứng minh: $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3$

Ngô Bá Hùng · Accepted Answer

$a+b+c+ab+bc+ca=6abc$

$\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$

Đặt $\left\{{}\begin{matrix}\frac{1}{a}=x\\frac{1}{b}=y\\frac{1}{c}=z\end{matrix}\right.$$\Rightarrow x+y+z+xy+yz+zx=6$

CM $P=x^2+y^2+z^2\ge3$

$x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z$

$2x^2+2y^2+2z^2\ge2xy+2yz+2zx$

Cộng vế với vế

$\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)=12$

$\Rightarrow3\left(x^2+y^2+z^2\right)\ge9$

$\Rightarrow x^2+y^2+z^2\ge3\left(đpcm\right)$

Vậy dấu "=" xảy ra khi $x=y=z=1$ hoặc $a=b=c=1$Câu trả lời: $a+b+c+ab+bc+ca=6abc$

$\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$

Đặt $\left\{{}\begin{matrix}\frac{1}{a}=x\\frac{1}{b}=y\\frac{1}{c}=z\end{matrix}\right.$$\Rightarrow x+y+z+xy+yz+zx=6$

CM $P=x^2+y^2+z^2\ge3$

$x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z$

$2x^2+2y^2+2z^2\ge2xy+2yz+2zx$

Cộng vế với vế

$\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)=12$

$\Rightarrow3\left(x^2+y^2+z^2\right)\ge9$

$\Rightarrow x^2+y^2+z^2\ge3\left(đpcm\right)$

Vậy dấu "=" xảy ra khi $x=y=z=1$ hoặc $a=b=c=1$

Nguyễn Việt Lâm · Answer

$a+b+c+ab+bc+ca=6abc$

$\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6$

Ta lại có:

$\frac{1}{a^2}+1+\frac{1}{b^2}+1+\frac{1}{c^2}+1-3\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}-3$

$\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\ge\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}$

Cộng vế với vế:

$\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)-3$

$\Leftrightarrow3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge6.2-3=9$

$\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3$

Dấu "=" xảy ra khi $a=b=c=1$Câu trả lời: $a+b+c+ab+bc+ca=6abc$