Đáp án: #Nguồn: doctailieu.com
Câu 5:
[tex]\frac{1}{a^2+b^2}+\frac{25}{ab}+ab=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{49}{2ab}+ab\geq \frac{4}{(a+b)^2}+\frac{32}{2ab}+ab+\frac{17}{2ab}\geq \frac{1}{4}+2\sqrt{\frac{32}{2ab}.ab}+\frac{17}{2ab}\geq \frac{1}{4}+2.4+\frac{17}{2.4}= \frac{83}{8}[/tex]
Dấu ''='' xảy ra khi: a=b=2