Các bạn giúp mình bài 5 với ạ mình cảm ơn
View attachment 133582
a, [tex]\frac{1}{ab}+\frac{1}{a^2+ab}+\frac{1}{ab+b^2}+\frac{1}{a^2+b^2}\\\\ =\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{1}{a^2+ab}+\frac{1}{ab+b^2}+\frac{1}{2ab}\\\\ \geq \frac{4}{(a+b)^2}+\frac{4}{(a+b)^2}+\frac{2}{(a+b)^2}\\\\ \geq 4+4+2=10[/tex]
dấu "=" <=> a=b=1/2
b, [tex]a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}\\\\ \geq \frac{(a+b)^2}{2}+\frac{(\frac{1}{a}+\frac{1}{b})^2}{2}=B\\\\ => B=\frac{16.(a+b)^2}{2}+\frac{(\frac{1}{a}+\frac{1}{b})^2}{2}-\frac{15.(a+b)^2}{2}\\\\ =\frac{(4a+4b)^2}{2}+\frac{(\frac{1}{a}+\frac{1}{b})^2}{2}-\frac{15(a+b)^2}{2}\\\\ \geq \frac{1}{2}.[\frac{(4a+\frac{1}{a}+4b+\frac{1}{b})^2}{2}]-\frac{15}{2}\\\\ \geq \frac{1}{4}.(4+4)^2-\frac{15}{2}=16-\frac{15}{2}=\frac{17}{2}[/tex]
dấu "=" <=> a=b=1/2
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
