Áp dụng BĐT Cauchy-Schwarz ta có:
\( P=\left(\frac{2}{a}+\frac{5}{b}+\frac{3}{c}\right)(a+b+c)\geq(\sqrt{2}+\sqrt{5}+\sqrt{3})^2\)
Vậy \(P_{Min}=\left(\sqrt{2}+\sqrt{5}+\sqrt{3}\right)^2\) khi
\(\left\{\begin{matrix}\frac{2}{a^2}=\frac{5}{b^2}=\frac{3}{c^2}\\\frac{2}{a}+\frac{5}{b}+\frac{3}{c}=1\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}a=2+\sqrt{6}+\sqrt{10}\\b=5+\sqrt{10}+\sqrt{15}\\c=3+\sqrt{6}+\sqrt{15}\end{matrix}\right.\)
bạn có biết pp nhân tử Lagrange là j k?