\(\frac{1}{a}+\frac{1}{b}=\frac{1}{2}\)
\(\Leftrightarrow2a+2b=ab\)
ta lại có:
\(\Delta_1=a^2-4b\)
\(\Delta_2=b^2-4a\)
\(\Rightarrow\Delta_1+\Delta_2=a^2-4b+b^2-4a=a^2+b^2-2.\left(2a+2b\right)\)
\(=a^2+b^2-2ab=\left(a-b\right)^2\ge0\)
\(\text{Vậy }\Delta_1+\Delta_2\ge0\)