Cho \(x,y>0\)và \(x+y\le2\)Tìm GTNN của \(A=x+y+\frac{2}{x}+\frac{2}{y}\)Mong các cao nhân hỗ trợ!
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\(\dfrac{3x^2}{2}+y^2+z^2+yz=1\)
\(\Leftrightarrow\dfrac{3}{2}x^2+\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}=1\)
Áp dụng BĐT Bunhiacopxki:
\(\left(\dfrac{2}{3}+1+\dfrac{1}{3}\right)\left(\dfrac{3}{2}x^2+\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}\right)\ge\left(\sqrt{\dfrac{2}{3}.\dfrac{3}{2}x^2}+\sqrt{1.\left(y+\dfrac{z}{2}\right)^2}+\sqrt{\dfrac{1}{3}.\dfrac{3z^2}{4}}\right)^2\)
\(\Leftrightarrow2.1\ge\left(x+y+\dfrac{z}{2}+\dfrac{z}{2}\right)^2=\left(x+y+z\right)^2\)
\(\Rightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)
\(\frac{3x^2}{2}+y^2+z^2+yz=1\)
\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)
\(\Rightarrow\left(x+y+z\right)^2\le2\)
\(\Leftrightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)
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\(\left(x+\frac{2}{x}\right)^2+\left(y+\frac{2}{y}\right)^2=x^2+y^2+\frac{4}{x^2}+\frac{4}{y^2}+4+4\)
\(=\left(x^2+\frac{1}{x^2}\right)+\left(y^2+\frac{1}{y^2}\right)+\left(\frac{3}{x^2}+3x+3x\right)+\left(\frac{3}{y^2}+3y+3y\right)-6\left(x+y\right)+8\)
\(\ge2+2+9+9-6.2+8=18\)
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1. \(1=x^2+y^2\ge2xy\Rightarrow xy\le\frac{1}{2}\)
\(A=-2+\frac{2}{1+xy}\ge-2+\frac{2}{1+\frac{1}{2}}=-\frac{2}{3}\)
max A = -2/3 khi x=y=\(\frac{\sqrt{2}}{2}\)
\(\frac{1}{xy}+\frac{1}{xz}=\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x}.\frac{4}{y+z}=\frac{4}{\left(4-t\right)t}=\frac{4}{4-\left(t-2\right)^2}\ge1\) với t = y+z => x =4 -t
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Áp dụng bđt Cô-si cho 2 số dương \(\frac{x}{2};\frac{8}{y}\) ta có:
\(\frac{x}{2}+\frac{8}{y}\ge2\sqrt{\frac{x}{2}\frac{8}{y}}=4\sqrt{\frac{x}{y}}\)
\(\Leftrightarrow2\ge4\sqrt{\frac{x}{y}}\Leftrightarrow0< \sqrt{\frac{x}{y}}\le\frac{1}{2}\Leftrightarrow0< \frac{x}{y}\le\frac{1}{4}\)
Đặt \(\frac{x}{y}=t\left(0< t\le\frac{1}{4}\right)\Rightarrow-t\ge\frac{-1}{4}\)
Ta có: \(K=t+\frac{2}{t}=32t+\frac{2}{t}-31t\ge2\sqrt{32t.\frac{2}{t}}-31t\ge16-\frac{31}{4}=\frac{33}{4}\)
Dấu '=' xảy ra <=> \(t=\frac{1}{4}\Leftrightarrow\hept{\begin{cases}x=2\\y=8\end{cases}}\)
Vậy GTNN của K là \(\frac{33}{4}\) tại x=2;y=8
\(2\ge\frac{x}{2}+\frac{8}{y}\ge2\sqrt{\frac{x}{2}.\frac{8}{y}}=4\sqrt{\frac{x}{y}}\Leftrightarrow\sqrt{\frac{x}{y}}\le\frac{1}{2}\Leftrightarrow\frac{y}{x}\ge4\)
\(K=\frac{x}{y}+\frac{2y}{x}=\frac{x}{y}+\frac{y}{16x}+\frac{31y}{16x}\ge2\sqrt{\frac{x}{y}.\frac{y}{16x}}+\frac{31}{16}.4=\frac{33}{4}\)
Dấu \(=\)xảy ra khi \(\hept{\begin{cases}\frac{x}{2}=\frac{y}{8}\\\frac{x}{2}+\frac{y}{8}=2\\\frac{x}{y}=\frac{y}{16x}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=8\end{cases}}\).
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\(A=x+\frac{1}{x}+y+\frac{1}{y}+\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{x}{x}}+2\sqrt{\frac{y}{y}}+\frac{4}{x+y}\ge2+2+\frac{4}{2}=6\)
\(A_{min}=6\) khi \(x=y=1\)
\(x+y\le2\Rightarrow-\left(x+y\right)\ge-2\)
Do đó:
\(A=2\left(x+\dfrac{1}{x}\right)+2\left(y+\dfrac{1}{y}\right)-\left(x+y\right)\ge2.2\sqrt{x.\dfrac{1}{x}}+2.2\sqrt{y.\dfrac{1}{y}}-2=6\)
\(A_{min}=6\) khi \(x=y=1\)