Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5}{16}\left(2x+y\right)\ge2\sqrt{\dfrac{3}{16}.3}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\).
Đẳng thức xảy ra khi x = 1; y = 2.
\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)
\(M=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5\left(2x+y\right)}{16}\ge2\sqrt{\dfrac{9\left(2x+y\right)}{16\left(2x+y\right)}}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{11}{4}\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;2\right)\)
Ta có:
\(M=\dfrac{2x+y}{xx}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)
\(=\left(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\right)+\dfrac{5}{8}\dfrac{2x+y}{2}\)
Có: \(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\ge2\sqrt{\dfrac{3}{8}\dfrac{2x+y}{2}\dfrac{3}{2x+y}}=\dfrac{3}{2}\)
Dấu '=' xảy ra \(\Leftrightarrow\dfrac{3}{8}\dfrac{2x+y}{2}=\dfrac{3}{2x+y}\)
Có: \(\dfrac{5}{8}\dfrac{2x+y}{2}\ge\dfrac{5}{8}\sqrt{2xy}=\dfrac{5}{4}\)
Dấu '=' xảy ra \(\Leftrightarrow2x=y,xy=2\)
\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\)
Dấu '=' xảy ra \(\Leftrightarrow x=1,y=2\)
Vậy GTNN của M là \(\dfrac{11}{4}\Leftrightarrow x=1,y=2\)
\(x\ge2y\Rightarrow\dfrac{x}{y}\ge2\)
\(M=\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{x}{4y}+\dfrac{y}{x}+\dfrac{3}{4}.\dfrac{x}{y}\ge2\sqrt{\dfrac{xy}{4xy}}+\dfrac{3}{4}.2=\dfrac{5}{2}\)
\(M_{min}=\dfrac{5}{2}\) khi \(x=2y\)
\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{1}{2xy}\)
Áp dụng BĐT Schwarz : \(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{\left(1+1\right)^2}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}=4\)
Lại có \(\dfrac{1}{2xy}=\dfrac{2}{4xy}\ge\dfrac{2}{\left(x+y\right)^2}=2\)
Cộng vế với vế được P \(\ge6\) ("=" khi x = y = 1/2)
Vậy Min P = 6 <=> x = y = 1/2