Áp dụng bđt Cauchy Schwarz ta có:
\(\left(3x+2y\right)^2\le\left(3^2+2^2\right)\left(x^2+y^2\right)=\left(9+4\right)\left(x^2+y^2\right)\)
\(\Leftrightarrow13^2\le13\left(x^2+y^2\right)\)\(\Leftrightarrow13\le x^2+y^2\)
Dấu "='' xảy ra khi x = 3; y = 2
\(3x+2y=13\Rightarrow y=\frac{13-3x}{2}\)
\(P=x^2+\frac{\left(13-3x\right)^2}{4}=\frac{4x^2+9x^2-3.13x+13^2}{4}=\frac{13}{4}\left(x^2-3x+13\right)\)
\(\frac{4P}{13}=x^2-3x+\frac{9}{4}+13-\frac{9}{4}=\left(x-\frac{3}{2}\right)^2+\frac{13.4-9}{4}\)
\(\frac{4P}{13}\ge\frac{13.4-9}{4}\Rightarrow P\ge\frac{13.\left(13.4-9\right)}{16}\)
Đẳng thức khi x=3/2
Làm lại cách cơ bạn sao lại sai được:
\(3x+2y=13\Rightarrow y=\frac{13-3x}{2}\)
\(P=x^2+\frac{\left(13-3x\right)^2}{4}=\frac{4x^2+9x^2-2.3.13x+13^2}{4}=\frac{13}{4}\left(x^2-2.3x+9+4\right)=\frac{13}{4}\left(x-3\right)^2+\frac{13}{4}.4\ge13\)GTLN(P)=13 khi x=3=> y=2
