Ta có: \(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
Lại có:
\(2=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\)
Mà \(a^2+b^2-ab>0\Rightarrow a+b>0\)
\(\Rightarrow2=\left(a+b\right)\left(a^2+b^2-ab\right)\ge\left(a+b\right).\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^3\)
\(\Rightarrow\left(a+b\right)^3\le8\Rightarrow a+b\le2\)
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