Đặt [tex]\sum _{cyc}\frac{a}{a^{2}+2b+3}=A[/tex]
Có: [tex]A=\sum _{cyc}\frac{a}{a^{2}+2b+3}= \sum _{cyc}\frac{a}{(a^{2}+1)+2b+2}\leq \sum _{cyc}\frac{a}{2a+2b+2}=\frac{1}{2}.\sum _{cyc}\frac{a}{a+b+1}[/tex]
[tex]\Rightarrow \frac{3}{2}-A\geq \frac{1}{2}\sum _{cyc}\frac{b+1}{a+b+1}[/tex]
Đặt $B= \frac{1}{2}\sum _{cyc}\frac{b+1}{a+b+1}$
Có: [tex]B=\sum _{cyc}\frac{b+1}{a+b+1}=\sum _{cyc}\frac{(b+1)^{2}}{(b+1)(a+b+1)}\geq \frac{(a+b+c+3)^{2}}{\sum_{cyc}[(b+1)(a+b+1)]}[/tex]
Đặt $C=\sum _{cyc}[(b+1)(a+b+1)]$
[tex]C=\sum _{cyc}[(b+1)(a+b+1)]\\=(a+1)(c+a+1)+(b+1)(a+b+1)+(c+1)(b+c+1)\\=a^{2}+b^{2}+c^{2}+ab+bc+ca+3(a+b+c)+3\\[/tex]
Có [tex]2C=2[a^{2}+b^{2}+c^{2}+ab+bc+ca+3(a+b+c)+3]\\=(a^2+b^2+c^2+2ab+2bc+2ca)+6(a+b+c)+6+a^2+b^2+c^2\\=(a+b+c)^2+6(a+b+c)+9\\=(a+b+c+3)^{2}[/tex]
Suy ra [tex]C=\frac{1}{2}(a+b+c+3)^{2}[/tex]
Khi đó:
[tex]B\geq \frac{(a+b+c+3)^{2}}{\sum_{cyc}[(b+1)(a+b+1)]}=\frac{(a+b+c+3)^{2}}{\frac{1}{2}(a+b+c+3)^{2}}=2[/tex]
Suy ra [tex]\frac{3}{2}-A\geq \frac{1}{2}.2=1\Leftrightarrow A\leq \frac{1}{2}[/tex]
Dấu = xảy ra khi [tex]a=b=c=1[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
