Ta có: \(2x+3y=7\Leftrightarrow\dfrac{x}{3}+\dfrac{y}{2}=\dfrac{7}{6}\)
\(3x^2+5y^2=\dfrac{\left(\dfrac{x}{3}\right)^2}{\dfrac{1}{27}}+\dfrac{\left(\dfrac{y}{2}\right)^2}{\dfrac{1}{20}}\ge\dfrac{\left(\dfrac{x}{3}+\dfrac{y}{2}\right)^2}{\dfrac{1}{27}+\dfrac{1}{20}}=\dfrac{\left(\dfrac{7}{6}\right)^2}{\dfrac{1}{27}+\dfrac{1}{20}}=\dfrac{735}{47}\) (đpcm)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\dfrac{70}{47}\\y=\dfrac{63}{47}\end{matrix}\right.\)