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\(H=x^2+2y^2+\frac{1}{x}+\frac{24}{y}=x^2+1+2y^2+8+\frac{1}{x}+\frac{24}{y}-9\)
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 \(\Leftrightarrow x=1;y=2\)
\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2xz}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{xz+2yz}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)
áp dụng BDT AM-GM \(=>x+y\ge2\sqrt{xy}=>\left(x+y\right)^2\ge4xy\left(1\right)\)
mà \(x+y\le1=>\left(x+y\right)^2\le1\left(2\right)\)
(1)(2)\(=>4xy\le\left(x+y\right)^2\le1=>4xy\le1=>xy\le\dfrac{1}{4}\)
\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\sqrt{1+x^2y^2}\ge2\sqrt{\dfrac{1+x^2y^2}{xy}}=2\sqrt{\dfrac{1}{xy}+xy}\)
\(=2\sqrt{\dfrac{1}{xy}+16xy-15xy}=2\sqrt{2\sqrt{16}-\dfrac{15}{4}}=\sqrt{17}\)
dấu"=" xảy ra<=>\(x=y=\dfrac{1}{2}\)
\(1\ge x+y\ge2\sqrt{xy}\Rightarrow xy\le\dfrac{1}{4}\Rightarrow\dfrac{1}{xy}\ge4\)
Ta có:
\(A\ge\dfrac{2}{\sqrt{xy}}.\sqrt{1+x^2y^2}=2\sqrt{\dfrac{1}{xy}+xy}=2\sqrt{\left(xy+\dfrac{1}{16xy}\right)+\dfrac{15}{16}.\dfrac{1}{xy}}\)
\(A\ge2\sqrt{2\sqrt{\dfrac{xy}{16xy}}+\dfrac{15}{16}.4}=\sqrt{17}\)
\(A_{min}=\sqrt{17}\) khi \(x=y=\dfrac{1}{2}\)
\(VT+3=\left(x+2y+3z+6\right)\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)\)
= \(24\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)\)
Áp dụng BĐT cauchy-schwarz:
\(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\ge\dfrac{9}{3+x+2y+3z}=\dfrac{9}{21}\)
\(\Rightarrow VT\ge\dfrac{24.9}{21}-3=\dfrac{51}{7}\)
dấu = xảy ra khi x=2y=3z=6 hay x=6,y=3,z=2
Theo đề thì:\(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{2}{z}=0\)
\(\Leftrightarrow xz+yz-2xy=0\)
Cũng từ \(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{2}{z}=0\)
\(\Leftrightarrow\dfrac{2}{z}=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\)
\(\Leftrightarrow z\le\sqrt{xy}\)
\(\Leftrightarrow z^2\le xy\)
Quay lại bài toán ta có:
\(T=\dfrac{x+z}{2x-z}+\dfrac{z+y}{2y-z}=\dfrac{2z^2-6xy-\left(xz+yz-2xy\right)}{-z^2+2\left(xz+yz-2xy\right)}\)
\(=\dfrac{6xy-2z^2}{z^2}\ge\dfrac{6xy-2xy}{xy}=4\)
Vậy GTNN là T = 4 khi x = y = z = 1
điểm rơi xấu quá: x=\(\dfrac{\sqrt[3]{9}}{2}\); y=\(\sqrt[3]{9}\), z =\(2\sqrt[3]{9}\) (4x=2y=z)
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 ....................................................................................................................