1. Cho a, b > 0 thỏa mãn : [tex]a + b \geq 2[/tex], tìm max của [tex]M = \frac{1}{a + b^2} + \frac{1}{b + a^2}[/tex]
[tex]M=\dfrac{1}{a+b^2}+\frac{1}{a^2+b} = \dfrac{a+1}{(a+b^2)(a+1)} + \dfrac{1+b}{(a^2+b)(1+b)}\leq \frac{a+1}{(a+b)^2}+\frac{b+1}{(a+b)^2}=\frac{a+b+2}{(a+b)^2}=\frac{1}{a+b}+\frac{2}{(a+b)^2}\leq \frac{1}{2}+\frac{2}{2^2}=1[/tex]
Last edited: