\(\sqrt{a}+\sqrt{b}=1\Rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2=1\)
\(\Rightarrow a+b+2\sqrt{ab}=1\)
\(\Rightarrow1-2\sqrt{ab}=a+b\)
Ta có
\(\left(4\sqrt{ab}+1\right)^2\ge0\)
\(\Rightarrow16ab-8\sqrt{ab}+1\ge0\)
\(\Rightarrow8\sqrt{ab}\left(1+2\sqrt{ab}\right)\le1\)
\(\Rightarrow8\sqrt{ab}\left(a+b\right)\le1\)
\(\Rightarrow64ab\left(a+b\right)\le1\)
\(\Rightarrow ab\left(a+b\right)\le\frac{1}{64}\)
(đpcm)