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ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge-3\end{matrix}\right.\) \(\Rightarrow x+y+1\ge0\)
Bình phương 2 vế giả thiết:
\(\left(x+y+1\right)^2=4\left(x+y+1+2\sqrt{\left(x-2\right)\left(y+3\right)}\right)\)
\(\Rightarrow\left(x+y+1\right)^2\le4\left(x+y+1+x+y+1\right)=5\left(x+y+1\right)\)
\(\Rightarrow x+y+1\le5\Rightarrow x+y\le4\)
Mặt khác:
\(\left(x+y+1\right)^2=4\left(x+y+1+2\sqrt{\left(x-2\right)\left(y+3\right)}\right)\ge4\left(x+y+1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+1\ge4\\x+y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y\ge3\\x+y=-1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}3\le S\le4\\S=-1\end{matrix}\right.\)
Ta có: \(x+y\ge2\sqrt{xy}\Rightarrow3xy\ge2\sqrt{xy}+1\Rightarrow3xy-2\sqrt{xy}-1\ge0\)
\(\Rightarrow\left(3\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)\ge0\Rightarrow\sqrt{xy}-1\ge0\) (do \(3\sqrt{xy}+1>0\) )
\(\Rightarrow\sqrt{xy}\ge1\Rightarrow xy\ge1\Rightarrow1-xy\le0\)
\(P=\dfrac{y\left(x+1\right)+x\left(y+1\right)}{xy\left(x+1\right)\left(y+1\right)}=\dfrac{2xy+x+y}{xy\left(xy+x+y+1\right)}\)
\(\Rightarrow P=\dfrac{2xy+3xy-1}{xy\left(xy+3xy\right)}=\dfrac{5xy-1}{4\left(xy\right)^2}=\dfrac{-4\left(xy\right)^2+5xy-1}{4\left(xy\right)^2}+1\)
\(\Rightarrow P=\dfrac{\left(1-xy\right)\left(4xy+1\right)}{4\left(xy\right)^2}+1\)
Do \(\left\{{}\begin{matrix}1-xy\le0\\4xy+1>0\\4\left(xy\right)^2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\left(1-xy\right)\left(4xy+1\right)}{4\left(xy\right)^2}\le0\)
\(\Rightarrow P\le0+1=1\Rightarrow P_{max}=1\) khi \(x=y=1\)
\(\left(x^2+\dfrac{8}{27x}+\dfrac{8}{27x}\right)+\left(y^2+\dfrac{8}{27y}+\dfrac{8}{27y}\right)+\dfrac{11}{27}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\ge3\sqrt[3]{\dfrac{8^2}{27^2}}+3\sqrt[3]{\dfrac{8^2}{27^2}}+\dfrac{11}{27}.\dfrac{4}{x+y}\)
\(\ge\dfrac{4}{3}+\dfrac{4}{3}+\dfrac{11}{9}=\dfrac{35}{9}\)
1/
\(S=\dfrac{1}{x}+\dfrac{2^2}{y}\ge\dfrac{\left(1+2\right)^2}{x+y}=\dfrac{9}{1}=9\)
\(\Rightarrow S_{min}=9\) khi \(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{2}{y}\\x+y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)
2/
Áp dụng BĐT: \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\Rightarrow x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Rightarrow\dfrac{\left(x+y\right)^2}{2}-3\left(x+y\right)\le x^2+y^2-3\left(x+y\right)=-4\)
\(\Rightarrow\dfrac{\left(x+y\right)^2}{2}-3\left(x+y\right)+4\le0\Leftrightarrow\left(x+y\right)^2-6\left(x+y\right)+8\le0\)
Đặt \(x+y=a\Rightarrow a^2-6a+8\le0\Rightarrow2\le a\le4\)
\(\Rightarrow2\le x+y\le4\)
\(\Rightarrow S\in\left[2;4\right]\)
thank you very much