Câu hỏi của Đặng Phương Nga

Question

Cho a,c,b dương t/m a+b+c+ab+bc+ac = 6abcCMR : $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3$

Kudo Shinichi · Accepted Answer

$a+b+c+ab+bc+ca=6abc\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6$Đặt $A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$Ta có : $\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}$CMTT :  $\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc};\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2.}{ca}$Ta có : $\left(\frac{1}{a}-1\right)^2\ge0\Leftrightarrow\frac{1}{a^2}+1\ge\frac{2}{a}$CMTT : $\frac{1}{b^2}+1\ge\frac{2}{b};\frac{1}{c^2}+1\ge\frac{2}{c}$$3A+3\ge2.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2.6=12$$\Leftrightarrow A+1\ge4\Leftrightarrow A\ge3\left(đpcm\right)$Chúc bạn học tốt !!!Câu trả lời: $a+b+c+ab+bc+ca=6abc\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6$Đặt $A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$Ta có : $\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}$CMTT :  $\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc};\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2.}{ca}$Ta có : $\left(\frac{1}{a}-1\right)^2\ge0\Leftrightarrow\frac{1}{a^2}+1\ge\frac{2}{a}$CMTT : $\frac{1}{b^2}+1\ge\frac{2}{b};\frac{1}{c^2}+1\ge\frac{2}{c}$$3A+3\ge2.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2.6=12$$\Leftrightarrow A+1\ge4\Leftrightarrow A\ge3\left(đpcm\right)$Chúc bạn học tốt !!!

Jack Dawson · Answer

$a+b+c+ab+ac+bc=6abc$$\Leftrightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$Đặt $\hept{\begin{cases}\frac{1}{a}=x\\frac{1}{b}=y\\frac{1}{c}=z\end{cases}}$ $\Rightarrow x+y+z+xy+xz+yz=6$Cần chứng minh $P=x^2+y^2+z^2\ge3$Ta có BĐT quen thuộc : $x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z$$2x^2+2y^2+2z^2\ge2xy+2xz+2yz$Cộng vế với vế : $\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+xz+yz\right)=12$ $\Rightarrow3\left(x^2+y^2+z^2\right)\ge9$$\Rightarrow x^2+y^2+z^2\ge3\left(đpcm\right)$Dấu " = " xảy ra khi $x=y=z=1$ hay $a=b=c=1$Câu trả lời: $a+b+c+ab+ac+bc=6abc$$\Leftrightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$Đặt $\hept{\begin{cases}\frac{1}{a}=x\\frac{1}{b}=y\\frac{1}{c}=z\end{cases}}$ $\Rightarrow x+y+z+xy+xz+yz=6$Cần chứng minh $P=x^2+y^2+z^2\ge3$Ta có BĐT quen thuộc : $x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z$$2x^2+2y^2+2z^2\ge2xy+2xz+2yz$Cộng vế với vế : $\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+xz+yz\right)=12$ $\Rightarrow3\left(x^2+y^2+z^2\right)\ge9$$\Rightarrow x^2+y^2+z^2\ge3\left(đpcm\right)$Dấu " = " xảy ra khi $x=y=z=1$ hay $a=b=c=1$

Bùi Anh Tuấn · Answer

Từ $a+b+c+ab+bc+ac=6\left(1\right)$Vì a,b,c dương nên ta chia hai vế của pt (1) cho abc ta có$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=6$Ta có$\frac{1}{a^2}+1\ge\frac{2}{a}$$\frac{1}{b^2}+1\ge\frac{2}{b}$$\frac{1}{c^2}+1\ge\frac{2}{c}$Và$\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}$$\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}$$\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}$Công theo BĐT ta có$3\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\right)\ge2\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)$$\Leftrightarrow3\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\right)\ge12$$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\ge4$$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3$Dấu $"="$xảy ra $\Leftrightarrow a+b+c=1$Câu trả lời: Từ $a+b+c+ab+bc+ac=6\left(1\right)$Vì a,b,c dương nên ta chia hai vế của pt (1) cho abc ta có$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=6$Ta có$\frac{1}{a^2}+1\ge\frac{2}{a}$$\frac{1}{b^2}+1\ge\frac{2}{b}$$\frac{1}{c^2}+1\ge\frac{2}{c}$Và$\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}$$\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}$$\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}$Công theo BĐT ta có$3\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\right)\ge2\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)$$\Leftrightarrow3\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\right)\ge12$$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\ge4$$\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3$Dấu $"="$xảy ra $\Leftrightarrow a+b+c=1$

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