Câu hỏi của Đặng Thanh Mai

Question

Bài 1 : Với a;b;c là những số thực thỏa mãn: ab+bc+ac=abc+a+b+c

với điều kiện  $3+ab\ne2;3+bc\ne2b+c;3+ac\ne2c+a$

CMR : \(\frac{1}{3+ab-\left(2a+b\right)}+\frac{1}{3+bc-\left(2b+c\right)}+\frac{1}...

Hoàng Thị Ánh Phương · Accepted Answer

Bài 1 :

Ta có : $ab+bc+ac=abc+a+b+c$

$\Leftrightarrow ab-abc+bc-b+ac-a-c=0$

$\Leftrightarrow ab-abc+bc-b+ac-a+1-c=1$

$\Leftrightarrow ab\left(1-c\right)+b\left(c-1\right)+a\left(c-1\right)+\left(1-c\right)=1$

$\Leftrightarrow ab\left(1-c\right)-b\left(1-c\right)-a\left(1-c\right)+\left(1-c\right)=1$

$\Leftrightarrow\left(1-c\right)\left(ab-a-b+1\right)=1$

$\Leftrightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)=1$

Ta có thế đặt $x=1-a;y=1-b;z=1-c\Rightarrow xyz=1$

Nhưng trong đẳng thức cần chứng minh theo $x;y;z$

$\Rightarrow$ Thế $a=1-x;b=1-y;c=1-z$ vào được :

$\frac{1}{3+ab-\left(2a+b\right)}=\frac{1}{3+\left(1-x\right)\left(1-y\right)-2\left(1-x\right)-\left(1-y\right)}=\frac{1}{1+x+xy}$

Tương tự :

$\frac{1}{3+ab-\left(2b+c\right)}=\frac{1}{3+\left(1-y\right)\left(1-z\right)-2\left(1-y\right)-\left(1-z\right)}=\frac{1}{1+y+yz}$

$\frac{1}{3+ac-\left(2c+a\right)}=\frac{1}{3+\left(1-x\right)\left(1-z\right)-2\left(1-z\right)-\left(1-x\right)}=\frac{1}{1+z+zx}$

Theo gt ta có xyz =1

$\Rightarrow VT=\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}$

$=\frac{1}{1+x+xy}+\frac{x}{x+xy+xyz}+\frac{xy}{xy+xyz+x^2yz}$

$=\frac{1}{1+x+xy}+\frac{x}{x+xy+1}+\frac{xy}{xy+1+x}$

$=\frac{1+x+xy}{1+x+xy}=1=VP$Câu trả lời: Bài 1 :

Ta có : $ab+bc+ac=abc+a+b+c$

$\Leftrightarrow ab-abc+bc-b+ac-a-c=0$

$\Leftrightarrow ab-abc+bc-b+ac-a+1-c=1$

$\Leftrightarrow ab\left(1-c\right)+b\left(c-1\right)+a\left(c-1\right)+\left(1-c\right)=1$

$\Leftrightarrow ab\left(1-c\right)-b\left(1-c\right)-a\left(1-c\right)+\left(1-c\right)=1$

$\Leftrightarrow\left(1-c\right)\left(ab-a-b+1\right)=1$

$\Leftrightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)=1$

Ta có thế đặt $x=1-a;y=1-b;z=1-c\Rightarrow xyz=1$

Nhưng trong đẳng thức cần chứng minh theo $x;y;z$

$\Rightarrow$ Thế $a=1-x;b=1-y;c=1-z$ vào được :

$\frac{1}{3+ab-\left(2a+b\right)}=\frac{1}{3+\left(1-x\right)\left(1-y\right)-2\left(1-x\right)-\left(1-y\right)}=\frac{1}{1+x+xy}$

Tương tự :

$\frac{1}{3+ab-\left(2b+c\right)}=\frac{1}{3+\left(1-y\right)\left(1-z\right)-2\left(1-y\right)-\left(1-z\right)}=\frac{1}{1+y+yz}$

$\frac{1}{3+ac-\left(2c+a\right)}=\frac{1}{3+\left(1-x\right)\left(1-z\right)-2\left(1-z\right)-\left(1-x\right)}=\frac{1}{1+z+zx}$

Theo gt ta có xyz =1

$\Rightarrow VT=\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}$

$=\frac{1}{1+x+xy}+\frac{x}{x+xy+xyz}+\frac{xy}{xy+xyz+x^2yz}$

$=\frac{1}{1+x+xy}+\frac{x}{x+xy+1}+\frac{xy}{xy+1+x}$

$=\frac{1+x+xy}{1+x+xy}=1=VP$

Hoàng Thị Ánh Phương · Answer

Bài 2 :

Áp dụng BĐT AM - GM

Ta có : $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{3}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}$

$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\ge\frac{3\sqrt[3]{abc}}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}$

Cộng theo vế ta được :

$\frac{1}{a+1}+\frac{a}{a+1}+\frac{1}{b+1}+\frac{b}{b+1}+\frac{1}{c+1}+\frac{c}{c+1}\ge\frac{3+3\sqrt[3]{abc}}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}$

$\Leftrightarrow1+1+1\ge\frac{3\left(\sqrt[3]{abc}+1\right)}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}$

$\Leftrightarrow3\ge\frac{3\left(\sqrt[3]{abc}+1\right)}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}$

$\Leftrightarrow3\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge3\left(\sqrt[3]{abc}+1\right)$

$\Leftrightarrow\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\sqrt[3]{abc}+1$

$\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge\left(\sqrt[3]{abc}+1\right)^3$

Dấu " = " xảy ra $\Leftrightarrow a=b=c$

Chúc bạn học tốt !!Câu trả lời: Bài 2 :