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Ta xét BĐT phụ: \(1+x^3+y^3\ge xy\left(x+y+z\right)\)
\(x^3+y^3\ge xy\left(x+y\right)+xyz-1\)
\(x^3+y^3-xy\left(x+y\right)\ge0\)
\(\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\ge0\)
\(\left(x+y\right)\left(x-y\right)^2\ge0\)( Luôn đúng, vậy BĐT phụ đúng)
\(\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}=\sqrt{x+y+z}.\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3\sqrt[3]{xyz}}.\left(3\sqrt[3]{\dfrac{1}{\sqrt{x^2y^2z^2}}}\right)=3\sqrt{3}\)
GTNN của P là \(3\sqrt{3}\Leftrightarrow x=y=z=1\)
+,3 = x + y + z \(\ge\) \(3\sqrt[3]{xyz}\Rightarrow xyz\le1\)
+, P \(\ge\) \(3\sqrt[3]{\dfrac{1}{xyz\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\ge\dfrac{3}{\sqrt[3]{\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\ge\dfrac{3}{\dfrac{x+y+z+3}{3}}=\dfrac{3}{2}\)
Áp dụng BĐT cauchy:
\(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\ge\dfrac{9}{xy+yz+zx}\)
\(M\ge\dfrac{1}{x^2+y^2+z^2}+\dfrac{9}{xy+yz+xz}=\dfrac{1}{x^2+y^2+z^2}+\dfrac{4}{2\left(xy+yz+xz\right)}+\dfrac{7}{xy+yz+zx}\)Áp dụng BĐT cauchy-schwarz:
\(\dfrac{1}{x^2+y^2+z^2}+\dfrac{4}{2\left(xy+yz+zx\right)}\ge\dfrac{\left(1+2\right)^2}{\left(x+y+z\right)^2}=9\)
và \(\dfrac{7}{xy+yz+xz}\ge\dfrac{7}{\dfrac{1}{3}\left(x+y+z\right)^2}=21\)
\(\Rightarrow M\ge9+21=30\)
dấu = xảy ra khi \(x=y=z=\dfrac{1}{3}\)
cô si cho đễ hiểu đi bn , cần gì phải cauchy s,. làm gì cho mệt
\(P=\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{z^2}{2}+\dfrac{x^2+y^2+z^2}{xyz}\)
\(\Rightarrow P\ge\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{z^2}{2}+\dfrac{xy+xz+yz}{xyz}\)
\(\Rightarrow P\ge\dfrac{x^2}{2}+\dfrac{1}{x}+\dfrac{y^2}{2}+\dfrac{1}{y}+\dfrac{z^2}{2}+\dfrac{1}{z}\)
\(\Rightarrow P\ge\left(\dfrac{x^2}{2}+\dfrac{1}{2x}+\dfrac{1}{2x}\right)+\left(\dfrac{y^2}{2}+\dfrac{1}{2y}+\dfrac{1}{2y}\right)+\left(\dfrac{z^2}{2}+\dfrac{1}{2z}+\dfrac{1}{2z}\right)\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{x^2}{2}.\dfrac{1}{2x}.\dfrac{1}{2x}}+3\sqrt[3]{\dfrac{y^2}{2}.\dfrac{1}{2y}.\dfrac{1}{2y}}+3\sqrt[3]{\dfrac{z^2}{2}.\dfrac{1}{2z}.\dfrac{1}{2z}}=\dfrac{9}{2}\)
\(\Rightarrow P_{min}=\dfrac{9}{2}\) khi \(x=y=z=1\)
Áp dụng BĐT Cauchy, ta có:
\(VT\ge3\sqrt[3]{\dfrac{x^2.y^2.z^2}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}=3\sqrt[3]{\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}\)
Ta có: xyz=1 và x,y,z >0
\(\Rightarrow x\le1\Rightarrow x+1\le2\Rightarrow\dfrac{1}{x+1}\ge\dfrac{1}{2}\)
Tương tự \(\dfrac{1}{y+1}\ge\dfrac{1}{2}\)
\(\dfrac{1}{z+1}\ge\dfrac{1}{2}\)
\(\Rightarrow VT\ge3\sqrt[3]{\dfrac{1}{x+1}.\dfrac{1}{y+1}.\dfrac{1}{z+1}}=\dfrac{3}{2}\)
Đẳng thức xảy ra khi x=y=z=1
Làm biến nghĩ nên làm cosi cho nó nhanh nhá:
Theo đề bài thì
\(3\sqrt[3]{xyz}\le x+y+z\le1\)
\(\Rightarrow xyz\le\dfrac{1}{27}\)
Ta có:
\(x+\dfrac{1}{y}=x+\dfrac{1}{9y}+\dfrac{1}{9y}+...+\dfrac{1}{9y}\ge10\sqrt[10]{\dfrac{x}{9^9y^9}}\left(1\right)\)
Tương tự ta có:
\(\left\{{}\begin{matrix}y+\dfrac{1}{z}\ge10\sqrt[10]{\dfrac{y}{9^9z^9}}\left(2\right)\\z+\dfrac{1}{x}\ge10\sqrt[10]{\dfrac{z}{9^9x^9}}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) ta có:
\(\Rightarrow\left(x+\dfrac{1}{y}\right)\left(y+\dfrac{1}{z}\right)\left(z+\dfrac{1}{x}\right)\ge1000\sqrt[10]{\dfrac{1}{9^{27}\left(xyz\right)^8}}=1000\sqrt[10]{\dfrac{27^8}{9^{27}}}=\dfrac{1000}{27}\)