Bất đẳng thức này >=3/2!!!!!!!!!!!!!

\(\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{a+b}+1-3=\left(a+b+c\right)\cdot\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)

áp dung cosy ta có   \(x+y+z\ge3\sqrt[3]{x\cdot y\cdot z}\)      \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{x\cdot y\cdot z}}\)

\(\Rightarrow\left(x+y+z\right)\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)

\(\Rightarrow\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge\frac{9}{2\cdot\left(a+b+c\right)}\)

\(\Rightarrow\left(a+b+c\right)\cdot\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\ge\frac{9}{2}\)

\(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{9}{2}-3=\frac{3}{2}\)

\(a^2+b^2\ge2ab\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow\left(a-b\right)^2\ge0\) (luôn đúng)

Vậy \(a^2+b^2\ge2ab\)

Áp dụng vào ta được :

\(a^2+1\ge2a\)

\(b^2+1\ge2b\)

\(c^2+1\ge2c\)

\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge2a.2b.2c=8abc\)(ĐPCM)

\(-\left(-a+b+c\right)+\left(b+c-1\right)=a-b-c+b+c-1=a-1\\ \left(b-c+6\right)-\left(7-a+b\right)=b-c+6-7+a-b=a-1\)

=> đpcm

Dùng phép khai triển. 

Không ai trả lời được à,câu này dễ mà