Ta có:[tex]\frac{a+b}{a+1}=\frac{(a+b)(a+1-a)}{a+1}=\frac{(a+b)(a+1)-a(a+b)}{a+1}=a+b-\frac{a(a+b)}{a+1}[/tex]
Lại có:[tex]\frac{1}{a+1}\leq \frac{a+1}{4a}\Rightarrow \frac{a+b}{a+1}\geq a+b-\frac{(a+b)(a+1)}{4}=\frac{4a+4b-a^2-ab-a-b}{4}=\frac{3(a+b)-a(a+b)}{4}=\frac{3}{4}(a+b)-\frac{1}{4}a^2-\frac{1}{4}ab[/tex]
[tex]\Rightarrow VT\geq \frac{3}{2}(a+b+c)-\frac{1}{4}(a^2+b^2+c^2)-\frac{1}{4}(ab+bc+ca)=\frac{9}{2}-\frac{1}{4}[(a+b+c)^2-2(ab+bc+ca)]-\frac{1}{4}(ab+bc+ca)=\frac{9}{2}-\frac{1}{4}.9+\frac{1}{2}(ab+bc+ca)-\frac{1}{4}(ab+bc+ca)=\frac{9}{4}+\frac{1}{4}(ab+bc+ca)[/tex]
Mà[tex](a+b+c)^2\geq 3(ab+bc+ca)\Rightarrow 9\geq 3(ab+bc+ca)\Rightarrow \frac{9}{4}\geq \frac{3}{4}(ab+bc+ca)\Rightarrow VT\geq \frac{3}{4}(ab+bc+ca)+\frac{1}{4}(ab+bc+ca)=ab+bc+ca[/tex]