Câu 5 ; Ta có P [tex]\geq \frac{1}{2(a+b+c)}-\frac{8}{a+b+c+5}\geq \frac{1}{2(a+b+c)}-\frac{1}{2(a+b+c)}-\frac{9}{10} =\frac{-9}{10}[/tex]
(do [tex]\sqrt{8bc}\leq 2c+b[/tex] ; [tex]\frac{16}{2(a+b+c)+10}\leq \frac{1}{2(a+b+c)}+\frac{9}{10}[/tex])
Dấu bằng [tex]a=c=\frac{5}{12},b=\frac{5}{6}[/tex]
Câu 5 ; Ta có P [tex]\geq \frac{1}{2(a+b+c)}-\frac{8}{a+b+c+5}\geq \frac{1}{2(a+b+c)}-\frac{1}{2(a+b+c)}-\frac{9}{10} =\frac{-9}{10}[/tex]
(do [tex]\sqrt{8bc}\leq 2c+b[/tex] ; [tex]\frac{16}{2(a+b+c)+10}\leq \frac{1}{2(a+b+c)}+\frac{9}{10}[/tex])
Dấu bằng [tex]a=c=\frac{5}{12},b=\frac{5}{6}[/tex]