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Thử nhé
Vì P là bất đẳng thức đối xứng nên dự đoán điểm rơi \(x=y=z=\dfrac{\sqrt{2021}}{3}\)
Thay vo P ta duoc \(P=4.\sqrt{2021}\)
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\(P=\sum\dfrac{\left(x+y\right)\sqrt{\left(y+z\right)\left(z+x\right)}}{z}\)
Cauchy-Schwarz:
\(\Rightarrow\left(y+z\right)\left(z+x\right)\ge\left(z+\sqrt{xy}\right)^2\Rightarrow\sqrt{\left(y+z\right)\left(z+x\right)}\ge z+\sqrt{xy}\)
\(\Rightarrow P\ge\sum\dfrac{\left(x+y\right)\left(z+\sqrt{xy}\right)}{z}\ge\sum\dfrac{xz+yz+x\sqrt{y}+y\sqrt{x}}{z}=\sum x+y+\dfrac{\left(x+y\right)\sqrt{xy}}{z}\ge\sum x+y+\dfrac{2xy}{z}\)
\(\Rightarrow P\ge2(x+y+z)+2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\)
Cauchy-Schwarz: \(\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\ge\left(\sqrt{\dfrac{xy}{z}.\dfrac{yz}{z}}+\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\right)^2=\left(x+y+z\right)^2\)
\(\Rightarrow P\ge2(x+y+z)+2\left(x+y+z\right)=4\left(x+y+z\right)=4\sqrt{2021}\)
\("="\Leftrightarrow x=y=z=\dfrac{\sqrt{2021}}{3}\)
\(A^2=\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}+2\left(\dfrac{xy}{\sqrt{yz}}+\dfrac{yz}{\sqrt{xz}}+\dfrac{xz}{\sqrt{xy}}\right)\)
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 \("="\Leftrightarrow x=y=z=\dfrac{12}{3}=4\)
\(P^2=\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}+\dfrac{2xy}{\sqrt{yz}}+\dfrac{2yz}{\sqrt{zx}}+\dfrac{2zx}{\sqrt{xy}}\)
\(P^2=\left(\dfrac{x^2}{y}+\dfrac{xy}{\sqrt{yz}}+\dfrac{xy}{\sqrt{yz}}+z\right)+\left(\dfrac{y^2}{z}+\dfrac{yz}{\sqrt{zx}}+\dfrac{yz}{\sqrt{zx}}+x\right)+\left(\dfrac{z^2}{x}+\dfrac{zx}{\sqrt{xy}}+\dfrac{zx}{\sqrt{xy}}+y\right)-\left(x+y+z\right)\)
\(P^2\ge4\sqrt[4]{\dfrac{x^4y^2z}{y^2z}}+4\sqrt[4]{\dfrac{y^4z^2x}{z^2x}}+4\sqrt[4]{\dfrac{z^4x^2y}{x^2y}}-\left(x+y+z\right)=3\left(x+y+z\right)\ge36\)
\(\Rightarrow P\ge6\)
\(P_{min}=6\) khi \(x=y=z=4\)
Áp dụng BĐT Cauchy:
\(\sqrt{\dfrac{x}{y+z}}+\sqrt{\dfrac{y}{z+x}}+\sqrt{\dfrac{z}{x+y}}\)
\(=\dfrac{x}{\sqrt{x\left(y+z\right)}}+\dfrac{y}{\sqrt{y\left(z+x\right)}}+\dfrac{z}{\sqrt{z\left(x+y\right)}}\)
\(\ge\dfrac{x}{\dfrac{x+y+z}{2}}+\dfrac{y}{\dfrac{x+y+z}{2}}+\dfrac{z}{\dfrac{x+y+z}{2}}\)
\(=\dfrac{2x}{x+y+z}+\dfrac{2y}{x+y+z}+\dfrac{2z}{x+y+z}\)
\(=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Dấu "=" không xảy ra nên \(\sqrt{\dfrac{x}{y+z}}+\sqrt{\dfrac{y}{z+x}}+\sqrt{\dfrac{z}{x+y}}>2\)
Ta có: \(\sqrt{\left(x^2+\dfrac{1}{y^2}\right)\left(1+81\right)}\ge\sqrt{\left(x+\dfrac{9}{y}\right)^2}\)
=> \(\sqrt{x^2+\dfrac{1}{y^2}}\ge\dfrac{x+\dfrac{9}{y}}{\sqrt{82}}\)
Tương tự => \(\left\{{}\begin{matrix}\sqrt{y^2+\dfrac{1}{z^2}}\ge\dfrac{y+\dfrac{9}{z}}{\sqrt{82}}\\\sqrt{z^2+\dfrac{1}{x^2}}\ge\dfrac{z+\dfrac{9}{x}}{\sqrt{82}}\end{matrix}\right.\)
=> \(P\ge\dfrac{\left(x+y+z\right)+9\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}{\sqrt{82}}\)
Mà x + y + z = 1
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}=9\)
=> \(P\ge\sqrt{82}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Ta có \(P^2=\left(\sum\dfrac{x}{\sqrt{y}}\right)^2=\sum\dfrac{x^2}{y}+2\left(\sum\dfrac{xy}{\sqrt{yz}}\right)\)
Mà \(\dfrac{x^2}{y}+\dfrac{xy}{\sqrt{yz}}+\dfrac{xy}{\sqrt{yz}}+z\ge4\sqrt[4]{x^4}=4x\)
Tương tự rồi cộng lại, ta có
\(P^2+x+y+z\ge4\left(x+y+z\right)\Rightarrow P^2\ge3\left(x+y+z\right)=36\Rightarrow P\ge6\)
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