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\(x\ge2y\Rightarrow x-y\ge y\Rightarrow x\left(x-y\right)\ge2y^2\Rightarrow x^2-xy-2y^2\ge0\).
\(\left(x-2y\right)^2\ge0\Leftrightarrow x^2-4xy+4y^2\ge0\)
\(\Rightarrow\left(x^2-xy-2y^2\right)+\left(x^2-4xy+4y^2\right)\ge0\)
\(\Leftrightarrow x^2+y^2\ge\frac{5}{2}xy\)
\(A=\frac{x^2+y^2}{xy}\ge\frac{\frac{5}{2}xy}{xy}=\frac{5}{2}\)
Dấu \(=\)xảy ra khi \(x=2y>0\).
Áp dụng bất đẳng thức AM - GM:
\(P=4x+3y+\dfrac{6}{x}+\dfrac{9}{2y}\)
\(=\left(\dfrac{3}{2}x+\dfrac{6}{x}\right)+\left(\dfrac{1}{2}y+\dfrac{9}{2y}\right)+\left(\dfrac{5}{2}x+\dfrac{5}{2}y\right)\)
\(\ge2\sqrt{\dfrac{3}{2}x\times\dfrac{6}{x}}+2\sqrt{\dfrac{1}{2}y\times\dfrac{9}{2y}}+\dfrac{5}{2}\times5\)
\(=\dfrac{43}{2}\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3}{2}x=\dfrac{6}{x}\\\dfrac{1}{2}y=\dfrac{9}{2y}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\left(\text{nhận}\right)\)
Vậy \(Min_P=\dfrac{43}{2}\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\)
Đặt \(\dfrac{x}{y}=a\Rightarrow0< a\le\dfrac{1}{4}\)
\(P=\dfrac{\left(\dfrac{x}{y}\right)^2-\dfrac{2x}{y}+2}{\dfrac{x}{y}+1}=\dfrac{a^2-2a+2}{a+1}=\dfrac{4a^2-8a+8}{4\left(a+1\right)}=\dfrac{4a^2-13a+3+5\left(a+1\right)}{4\left(a+1\right)}\)
\(P=\dfrac{5}{4}+\dfrac{\left(1-4a\right)\left(3-a\right)}{4\left(a+1\right)}\ge\dfrac{5}{4}\)
Dấu "=" xảy ra khi \(a=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)
Áp dụng bđt \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\)
\(x^2+2y^2\ge\dfrac{\left(x+2y\right)^2}{2}=\dfrac{25}{2}\)
Ta có:
\(x+2y\ge2\sqrt{x2y}\)
\(\Leftrightarrow5\ge2\sqrt{2xy}\)
\(\Rightarrow25\ge4.2xy\Rightarrow xy\le\dfrac{25}{8}\)
Áp dụng bđt Cosi
\(\dfrac{1}{x}+\dfrac{24}{y}\ge2\sqrt{\dfrac{24}{xy}}\ge2\sqrt{\dfrac{24}{\dfrac{25}{8}}}=2\sqrt{\dfrac{24.8}{25}}=\dfrac{16}{5}\sqrt{3}\)
\(\Rightarrow H\ge\dfrac{16}{5}\sqrt{3}+\dfrac{25}{2}\)
Dấu bằng xảy ra khi:
\(\left\{{}\begin{matrix}x=2y\\x+2y=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{5}{4}\end{matrix}\right.\)
ta có : \(H=x^2+2y^2+\dfrac{1}{x}+\dfrac{24}{y}=x^2+\dfrac{1}{x}+2y^2+\dfrac{24}{y}\)
\(\Rightarrow H\ge2\sqrt{x}+2\sqrt{48y}\) dấu "=" xảy ra khi \(x=1;y=2\)
thế lại ta có : \(H_{min}=2+8\sqrt{6}\)
vậy ....................................................................................................................
\(H=x^2+2y^2+\frac{1}{x}+\frac{24}{y}=x^2+1+2y^2+8+\frac{1}{x}+\frac{24}{y}-9\)
Vì x ; y > 0 , áp dụng BĐT Cauchy , ta có :
\(H\ge2x+8y+\frac{1}{x}+\frac{24}{y}-9=x+2y+x+\frac{1}{x}+6\left(y+\frac{4}{y}\right)-9\)
\(\ge5+2+6.4-9=22\)
Dấu " = " xảy ra \(\Leftrightarrow x=1;y=2\)