Ta có: [tex]\frac{1}{k^2+1}-\frac{a}{(k^2+1)a+k(b+c)}=\frac{k(b+c)}{(k^2+1)[(k^2+1)a+(b+c)]}=\frac{k}{k+1}.\frac{b+c}{(k^2+1)a+(b+c)}[/tex]
Áp dụng BĐT Cauchy - Schwartz ta có: [tex]\frac{b+c}{(k^2+1)a+k(b+c)}+\frac{c+a}{(k^2+1)b+k(c+a)}+\frac{a+b}{(k^2+1)c+k(a+b)}=\frac{(b+c)^2}{(k^2+1)a(b+c)+k(b+c)^2}+\frac{(c+a)^2}{(k^2+1)b(c+a)+k(c+a)^2}+\frac{(a+b)^2}{(k^2+1)c(a+b)+k(a+b)^2} \geq \frac{[2(a+b+c)]^2}{2(k^2+1)(ab+bc+ca)+k[(a+b)^2+(b+c)^2+(c+a)^2]}=\frac{4(a+b+c)^2}{2(k^2+k+1)(ab+bc+ca)+2k(a^2+b^2+c^2)}=\frac{2(a+b+c)^2}{(k^2+k+1)(ab+bc+ca)+k(a^2+b^2+c^2)}=\frac{2(a+b+c)^2}{k(a+b+c)^2+(k^2-k+1)(ab+bc+ca)} \geq \frac{2(a+b+c)^2}{k(a+b+c)^2+\frac{k^2-k+1}{3}(a+b+c)^2}=\frac{6}{(k+1)^2}[/tex]
Từ đó [tex]\frac{3}{k^2+1}-\frac{a}{(k^2+1)a+k(b+c)}-\frac{b}{(k^2+1)b+k(a+c)}-\frac{c}{(k^2+1)c+k(a+b)} \geq \frac{k}{k^2+1}.\frac{6}{(k+1)^2}=\frac{6k}{(k+1)^2(k^2+1)}\Rightarrow \frac{a}{(k^2+1)a+k(b+c)}+\frac{b}{(k^2+1)b+k(a+c)}+\frac{c}{(k^2+1)c+k(a+b)} \leq \frac{3}{k^2+1}-\frac{6k}{(k^2+1)(k+1)^2}=\frac{3}{(k+1)^2}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
