c) Áp dụng Bđt Cauchy cho 2 số dương \(3a;5b\)
\(3a+5b\ge2\sqrt{3a.5b}\)
\(\Leftrightarrow12\ge2\sqrt{15}.\sqrt{ab}\)
\(\Leftrightarrow144\ge60ab\)
\(\Leftrightarrow P=ab\le\dfrac{144}{60}=\dfrac{12}{5}\)
Dấu "=" xảy ta khi \(3a=5b\)
\(\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{3}=\dfrac{3a+5b}{3.5+5.3}=\dfrac{12}{30}=\dfrac{2}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=\dfrac{6}{5}\end{matrix}\right.\)
Vậy \(GTLN\left(P\right)=\dfrac{12}{5}\left(tại.a=2;b=\dfrac{6}{5}\right)\)
a) Ta có :
\(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0,\forall a;b\ge0\)
\(\Leftrightarrow a+b-2\sqrt{ab}\)
\(\Leftrightarrow a+b\ge2\sqrt{ab}\left(a;b\ge0\right)\)
\(\Rightarrowđpcm\)
Câu b chứng minh gì vậy bạn?
