V
vuanoidoi
[tex](a-1)(a-2) \le 0 =>a^2-3a+2 \le 0 =>a^2 \le 3a-2 => a \le 3- \frac{2}{a} =>( a+b+c)( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ) \ge (9-\frac{2}{a} + \frac{2}{b} + \frac{2}{c})(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) =9A-2A^2 \le 10 [/tex]2.
Tìm max :
[TEX]M = ( a+b+c)( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ) , \ \ \forall a,b,c \in [1;2][/TEX]
Cho x,y là 2 số thực bất kì khác 0
Min???? : [TEX]\frac{4x^2y^2}{(x^2+y^2)^8}+ \frac{x^2}{y^2}+ \frac{y^2}{x^2}[/TEX]
[TEX]\frac{4x^2y^2}{(x^2+y^2)^8}+ \frac{x^2}{y^2}+ \frac{y^2}{x^2}= \frac{4ab}{(a+b)^8}+ \frac{a}{b}+ \frac{b}{a}=\frac{4ab}{(a+b)^8}+\frac{a^2+b^2}{ab} \ge \frac{4ab}{(a+b)^8}+\frac{(a+b)^2}{2ab} (AM-GM)[/TEX]
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