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đặt 2x+3=a
\(y\sqrt{y}+y=a\sqrt{a}+a\)
=>\(\left(\sqrt{y}-\sqrt{a}\right)\left(y+\sqrt{ay}+a+\sqrt{a}+\sqrt{y}\right)=0\)
=>\(\sqrt{y}=\sqrt{a}\Rightarrow y=2x+3\)
thay vào Q tìm min là xong
Đặt \(\hept{\begin{cases}\sqrt{2x+3}=a\left(a>0\right)\\\sqrt{y}=b\left(b\ge0\right)\end{cases}}\)
Thì ta có
\(\frac{b^2}{a^2}=\frac{a+1}{b+1}\)
\(\Leftrightarrow b^3+b^2=a^3+a^2\)
\(\Leftrightarrow\left(b-a\right)\left(b^2+ab+a^2\right)+\left(b-a\right)\left(b+a\right)=0\)
\(\Leftrightarrow\left(b-a\right)\left(b^2+ab+a^2+b+a\right)=0\)
Mà \(\left(b^2+ab+a^2+b+a\right)>0\)
\(\Rightarrow a=b\)
\(\Rightarrow2x+3=y\)
Thế vào Q ta được
\(Q=2x^2-5x-12=\left(2x^2-\frac{2x\times\sqrt{2}\times5}{2\sqrt{2}}+\frac{25}{8}\right)-\frac{121}{8}\)
\(=\left(\sqrt{2}x-\frac{5}{2\sqrt{2}}\right)^2-\frac{121}{8}\ge\frac{-121}{8}\)
Ta có: \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac52\)
=>\(\frac{x}{\sqrt{xy}}+\frac{y}{\sqrt{xy}}=\frac52\)
=>\(2\left(x+y\right)=5\sqrt{xy}\)
=>\(2x-5\sqrt{xy}+2y=0\)
=>\(2x-4\sqrt{xy}-\sqrt{xy}+2y=0\)
=>\(2\sqrt{x}\left(\sqrt{x}-2\sqrt{y}\right)-\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)=0\)
=>\(\left(\sqrt{x}-2\sqrt{y}\right)\left(2\sqrt{x}-\sqrt{y}\right)=0\)
TH1: \(\sqrt{x}-2\sqrt{y}=0\)
=>\(\sqrt{x}=2\sqrt{y}\)
=>x=4y
\(A=\frac{2x+3\cdot\sqrt{xy}}{2x-3\sqrt{xy}}\)
\(=\frac{\sqrt{x}\left(2\sqrt{x}+3\sqrt{y}\right)}{\sqrt{x}\left(2\sqrt{x}-3\sqrt{y}\right)}=\frac{2\sqrt{x}+3\sqrt{y}}{2\sqrt{x}-3\sqrt{y}}\)
\(=\frac{2\cdot\sqrt{4y}+3\sqrt{y}}{2\cdot\sqrt{4y}-3\sqrt{y}}=\frac{4\sqrt{y}+3\sqrt{y}}{4\sqrt{y}-3\sqrt{y}}=\frac{4+3}{4-3}=7\)
TH2: \(2\sqrt{x}-\sqrt{y}=0\)
=>\(\sqrt{y}=2\sqrt{x}\)
=>y=4x
\(A=\frac{2x+3\cdot\sqrt{xy}}{2x-3\sqrt{xy}}\)
\(=\frac{\sqrt{x}\left(2\sqrt{x}+3\sqrt{y}\right)}{\sqrt{x}\left(2\sqrt{x}-3\sqrt{y}\right)}=\frac{2\sqrt{x}+3\sqrt{y}}{2\sqrt{x}-3\sqrt{y}}\)
\(=\frac{2\sqrt{x}+3\cdot\sqrt{4x}}{2\sqrt{x}-3\cdot\sqrt{4x}}=\frac{2\sqrt{x}+6\sqrt{x}}{2\sqrt{x}-6\sqrt{x}}=\frac{8}{-4}=-2\)
\(P=\sqrt{\frac{1}{36}\left(11a+7b\right)^2+\frac{59\left(a-b\right)^2}{36}}+\sqrt{\frac{1}{36}\left(7a+11b\right)+\frac{59\left(a-b\right)^2}{36}}\)
\(=\sqrt{\frac{1}{16}\left(3a+5b\right)^2+\frac{5\left(a-b\right)^2}{16}}+\sqrt{\frac{1}{16}\left(5a+3b\right)^2+\frac{5\left(a-b\right)^2}{16}}\)
\(\ge\frac{1}{6}\left(11a+7b\right)+\frac{1}{6}\left(7a+11b\right)+\frac{1}{4}\left(3a+5b\right)+\frac{1}{4}\left(5a+3b\right)\)
\(=5\left(a+b\right)=5.2016=10080\)
Ta có BĐT:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(\Leftrightarrow6\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)+2016\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow7.\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le2016\)
Xét \(P=\frac{1}{\sqrt{3\left(2x^2+y^2\right)}}+\frac{1}{\sqrt{3\left(2y^2+z^2\right)}}+\frac{1}{\sqrt{3\left(2z^2+x^2\right)}}\)
\(P^2=\left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2x^2+y^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2y^2+z^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2z^2+x^2}}\right)^2\)
Áp dụng BĐT Bunhiacopxki ta có:
\(P^2\le\left(\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2\right)\left(\left(\frac{1}{\sqrt{2x^2+y^2}}\right)^2+\left(\frac{1}{\sqrt{2y^2+z^2}}\right)^2+\left(\frac{1}{\sqrt{2z^2+x^2}}\right)^2\right)\)
\(\Leftrightarrow P^2\le\frac{1}{2x^2+y^2}+\frac{1}{2y^2+z^2}+\frac{1}{2z^2+x^2}\)
Mặt khác ta có:
\(\frac{1}{2x^2+y^2}=\frac{1}{x^2+x^2+y^2}\le\frac{1}{9}\left(\frac{1}{x^2}+\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\frac{1}{2y^2+z^2}\le\frac{1}{9}\left(\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
\(\frac{1}{2z^2+x^2}\le\frac{1}{9}\left(\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}\right)\)
\(\Rightarrow P^2\le\frac{1}{3}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le\frac{1}{3}.2016=672\)
\(\Rightarrow P\le4\sqrt{42}\)
Dấu '=' xảy ra khi \(x=y=z=\sqrt{\frac{1}{672}}\)