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Áp dụng BĐT cosi:
\(A=\left(3x+\dfrac{3}{x}\right)+\left(\dfrac{4}{9}y+\dfrac{4}{y}\right)+\left(2x+y\right)\\ A\ge2\sqrt{\dfrac{9x}{x}}+2\sqrt{\dfrac{16y}{9y}}+5\\ A\ge2\cdot3+2\cdot\dfrac{4}{3}+5=\dfrac{41}{3}\)
Vậy \(A_{min}=\dfrac{41}{3}\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{3}{x}\\\dfrac{4y}{9}=\dfrac{4}{y}\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
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\(A^2=\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}+2\left(\dfrac{xy}{\sqrt{yz}}+\dfrac{yz}{\sqrt{xz}}+\dfrac{xz}{\sqrt{xy}}\right)\)
Áp dụng BĐT cosi:
\(\dfrac{x^2}{y}+\dfrac{xy}{\sqrt{yz}}+\dfrac{xy}{\sqrt{yz}}+z\ge4\sqrt[4]{\dfrac{x^4y^2z}{y^2z}}=4x\)
\(\dfrac{y^2}{z}+\dfrac{yz}{\sqrt{xz}}+\dfrac{yz}{\sqrt{xz}}+x\ge4\sqrt[4]{\dfrac{y^4z^2x}{z^2x}}=4y\)
\(\dfrac{z^2}{x}+\dfrac{xz}{\sqrt{xy}}+\dfrac{xz}{\sqrt{xy}}+y\ge4\sqrt[4]{\dfrac{z^4x^2y}{x^2z}}=4z\)
Cộng VTV 3 BĐT trên:
\(\Leftrightarrow A^2+\left(x+y+z\right)\ge4\left(x+y+z\right)\\ \Leftrightarrow A^2\ge3\left(x+y+z\right)\ge3\cdot12=36\\ \Leftrightarrow A\ge6\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{12}{3}=4\)
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\(\dfrac{\left(x+y+1\right)^2}{xy+x+y}\ge\dfrac{3\left(xy+x+y\right)}{xy+x+y}=3\)
\(\Rightarrow A=\dfrac{8\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{xy+x+y}{\left(x+y+1\right)^2}\)
\(A\ge\dfrac{8}{9}.3+2\sqrt{\dfrac{\left(x+y+1\right)^2\left(xy+x+y\right)}{\left(xy+x+y\right)\left(x+y+1\right)^2}}=\dfrac{10}{3}\)
Dấu "=" xảy ra khi \(x=y=1\)
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ÁP dụng BĐT Mincopxki, ta có:
\(A\ge\sqrt{\left(x+y\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}\)
\(=\sqrt{\left(x+y\right)^2+\dfrac{\left(x+y\right)^2}{\left(xy\right)^2}}\)
\(\ge\sqrt{2\sqrt{\left(x+y\right)^2.\dfrac{\left(x+y\right)^2}{\left(xy\right)^2}}}=\sqrt{\dfrac{2\left(x+y\right)^2}{xy}}\) (cô si)
\(\ge\sqrt{\dfrac{2.4xy}{xy}}=\sqrt{8}=2\sqrt{2}\left(Côsi\right)\)
Min \(A=2\sqrt{2}\Leftrightarrow x=y\)
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C1:
\(x,y>0\)
\(M=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2=x^2+2+\dfrac{1}{x^2}+y^2+2+\dfrac{1}{y^2}=\left(x^2+\dfrac{1}{16x^2}\right)+\left(y^2+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+4\)Theo BĐT AM-GM (Caushy) ta có:
\(M=\left(x^2+\dfrac{1}{16x^2}\right)+\left(y^2+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+4\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}.2\sqrt{\dfrac{1}{x^2}.\dfrac{1}{y^2}}+4=\dfrac{1}{2}+\dfrac{1}{2}+4+\dfrac{15}{4}.\dfrac{1}{xy}\ge5+\dfrac{15}{4}.\dfrac{1}{\left(\dfrac{x+y}{2}\right)^2}\ge5+\dfrac{15}{4}.\dfrac{1}{\left(\dfrac{1}{2}\right)^2}=20\)Đẳng thức xảy ra \(\left\{{}\begin{matrix}x^2=\dfrac{1}{16}x^2\\y^2=\dfrac{1}{16}y^2\\x+y=1\\x,y>0\end{matrix}\right.\Leftrightarrow x=y=\dfrac{1}{2}\)
Vậy \(MinM=20\)
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\(A\ge\dfrac{\left(x+y\right)^2}{2xy}+\dfrac{\sqrt{xy}}{x+y}\)
\(A\ge\dfrac{7\left(x+y\right)^2}{16xy}+\dfrac{\left(x+y\right)^2}{16xy}+\dfrac{\sqrt{xy}}{2\left(x+y\right)}+\dfrac{\sqrt{xy}}{2\left(x+y\right)}\)
\(A\ge\dfrac{7.4xy}{16xy}+3\sqrt[3]{\dfrac{\left(x+y\right)^2xy}{16.4.xy\left(x+y\right)^2}}=\dfrac{5}{2}\)
Dấu "=" xảy ra khi \(x=y\)
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Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)
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Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có
\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)
\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)
Lời giải:
$A=(x-y)+\frac{4}{x-y}+y+\frac{1}{y}$
Áp dụng BĐT Cô-si:
$(x-y)+\frac{4}{x-y}\geq 2\sqrt{(x-y).\frac{4}{x-y}}=4$
$y+\frac{1}{y}\geq 2$
$\Rightarrow A\geq 4+2=6$
Vậy $A_{\min}=6$ khi $(x,y)=(3,1)$