Áp dụng BĐT Cauchy-Schwarz dạng engel:
\(\dfrac{2}{x}+\dfrac{3}{y}=\dfrac{\sqrt{2}^2}{x}+\dfrac{\sqrt{3}^2}{y}\ge\dfrac{\left(\sqrt{2}+\sqrt{3}\right)^2}{x+y}\)
\(\Rightarrow\dfrac{\left(\sqrt{2}+\sqrt{3}\right)^2}{x+y}\le6\Rightarrow x+y\ge\dfrac{\left(\sqrt{2}+\sqrt{3}\right)^2}{6}=\dfrac{5+2\sqrt{6}}{6}\)
\(\Rightarrow\left(x+y\right)_{min}=\dfrac{5+2\sqrt{6}}{6}\)
Dấu "=" xảy ra khi \(\dfrac{\sqrt{2}}{x}=\dfrac{\sqrt{3}}{y}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+\sqrt{6}}{6}\\y=\dfrac{3+\sqrt{6}}{6}\end{matrix}\right.\)
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