Ta sẽ chứng minh [tex]P \geq 4 \Rightarrow ab+bc+ca+1 \geq 4abc \Rightarrow 2a+2b+2c+2 \geq 8abc[/tex]
[tex]\Rightarrow 4ab+4bc+4ca+4a+4b+4c+3 \geq 8abc+4ab+4bc+4ca+2a+2b+2c+1[/tex]
[tex]\Rightarrow (2a+1)(2b+1)+(2b+1)(2c+1)+(2c+1)(2a+1) \geq (2a+1)(2b+1)(2c+1)[/tex]
[tex]\Rightarrow \dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1} \geq 1[/tex]
Thật vậy:
[tex]VT=\sum\dfrac{b^2c^2}{2ab^2c^2+b^2c^2} \geq \dfrac{(ab+bc+ca)^2}{2abc(a+b+c)+a^2b^2+b^2c^2+c^2a^2}=\dfrac{(ab+bc+ca)^2}{(ab+bc+ca)^2}=1[/tex]