\(P=\dfrac{\sqrt{x}+1}{x\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(maxP=\dfrac{4}{3}\Leftrightarrow x=\dfrac{1}{4}\)
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