Theo BĐT cô-si ta có:
\(\dfrac{a+b}{2}\ge\sqrt{ab}\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2=\left(\dfrac{2}{2}\right)^2=1\)
\(\dfrac{a^2+b^2}{2}\ge\sqrt{a^2b^2}\Rightarrow a^2+b^2\ge2\sqrt{1^2}=2\)
Ta có:
\(\dfrac{a}{b+1}+\dfrac{b}{a+1}\)
\(=\dfrac{a\left(a+1\right)}{\left(b+1\right)\left(a+1\right)}+\dfrac{b\left(b+1\right)}{\left(b+1\right)\left(a+1\right)}\)
\(=\dfrac{a^2+a+b^2+b}{\left(b+1\right)\left(a+1\right)}\)
\(=\dfrac{\left(a^2+b^2\right)+\left(a+b\right)}{ab+a+b+1}\ge\dfrac{2+2}{1+2+1}=1\left(dpcm\right)\)
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