Lời giải:
Vì \(\left\{\begin{matrix} a+b=2m\\ ab=m^2\end{matrix}\right.\Rightarrow \left\{\begin{matrix} (a+b)^2=4m^2\\ 4ab=4m^2\end{matrix}\right.\)
\(\Rightarrow (a+b)^2=4ab\)
\(\Leftrightarrow a^2+2ab+b^2-4ab=0\)
\(\Leftrightarrow a^2-2ab+b^2=0\Leftrightarrow (a-b)^2=0\Rightarrow a=b\)
Ta có đpcm.
Cách khác
\(ab\le\dfrac{\left(a+b\right)^2}{4}\Leftrightarrow m^2\le\dfrac{\left(a+b\right)^2}{4}\Leftrightarrow m\le\dfrac{a+b}{2}\)
\(\Leftrightarrow2m\le a+b\). Theo đề \(2m=a+b\)
\("="\Leftrightarrow a=b\)