Ta có:
\(\dfrac{9}{4}=\left(1+a\right)\left(1+b\right)\le\dfrac{1}{4}\left(1+a+1+b\right)^2\)
\(\Rightarrow\left(a+b+2\right)^2\ge9\)
\(\Rightarrow a+b+2\ge3\)
\(\Rightarrow a+b\ge1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)
Lại có: với mọi a;b thì:
\(\left(a-\dfrac{1}{2}\right)^2+\left(b-\dfrac{1}{2}\right)^2\ge0\)
\(\Rightarrow a^2+b^2+\dfrac{1}{2}\ge a+b\ge1\)
\(\Rightarrow a^2+b^2\ge\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)
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