Đặt [imath]VT=A[/imath]
Áp dụng BĐT Cauchy dạng: [imath]\dfrac{1}{x+y} \leq \dfrac{1}{4}(\dfrac{1}{x}+\dfrac{1}{y})[/imath]
[imath]=>\dfrac{1}{ab+1+a+1} \leq \dfrac{1}{4}(\dfrac{1}{ab+1}+\dfrac{1}{a+1})[/imath]
[imath]=> \dfrac{1}{ab+a+2} \leq \dfrac{1}{4}(\dfrac{abc}{ab+abc}+\dfrac{1}{a+1})[/imath]
[imath]=> \dfrac{1}{ab+a+2} \leq \dfrac{1}{4}(\dfrac{c}{c+1}+\dfrac{1}{a+1})[/imath]
Tương tự: [imath]\dfrac{1}{bc+b+2} \leq \dfrac{1}{4}(\dfrac{a}{a+1}+\dfrac{1}{b+1})[/imath]
[imath]\dfrac{1}{ca+c+2} \leq \dfrac{1}{4}(\dfrac{b}{b+1}+\dfrac{1}{c+1})[/imath]
Cộng theo vế:
[imath]=> A \leq \dfrac{1}{4}.(1+1+1)[/imath]
[imath]=> A \leq \dfrac{3}{4}[/imath]
Dấu [imath]'='[/imath] xảy ra khi: [imath]a=b=c=1[/imath]