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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 \("="\Leftrightarrow x=y=z=\dfrac{12}{3}=4\)
TA có \(P^2=\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}+2\left(\dfrac{xy}{\sqrt{yz}}+\dfrac{yz}{\sqrt{zx}}+\dfrac{zx}{\sqrt{xy}}\right)\)
Áp dụng BĐt AM-GM, ta có \(\dfrac{x^2}{y}+\dfrac{xy}{\sqrt{yz}}+\dfrac{xy}{\sqrt{yz}}+z\ge4x\)
tương tự rồi cộng lại, ta có \(P^2+\left(x+y+z\right)\ge4\left(x+y+z\right)\Rightarrow P^2\ge3\left(x+y+z\right)\ge36\Rightarrow P\ge6\)
dấu = xảy ra <=> x=y=z=4
Đặt \(\left\{{}\begin{matrix}\sqrt{y+z-4}=a>0\\\sqrt{z+x-4}=b>0\\\sqrt{x+y-4}=c>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{b^2+c^2-a^2+4}{2}\\y=\dfrac{c^2+a^2-b^2+4}{2}\\z=\dfrac{a^2+b^2-c^2+4}{2}\end{matrix}\right.\).
\(2P=\dfrac{b^2+c^2-a^2+4}{a}+\dfrac{c^2+a^2-b^2+4}{b}+\dfrac{a^2+b^2-c^2+4}{c}=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}-a-b-c\).
Áp dụng bất đẳng thức AM - GM:
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}=\left(\dfrac{a^2}{b}+b\right)+\left(\dfrac{b^2}{c}+c\right)+\left(\dfrac{c^2}{a}+a\right)-\left(a+b+c\right)\ge2a+2b+2c-a-b-c=a+b+c\).
Tương tự, \(\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}\ge a+b+c\).
Do đó \(2P\ge a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}=\left(a+\dfrac{4}{a}\right)+\left(b+\dfrac{4}{b}\right)+\left(c+\dfrac{4}{c}\right)\ge4+4+4=12\Rightarrow P\ge6\).
Đẳng thức xảy ra khi a = b = c = 2 hay x = y = z = 4.
Vậy Min P = 6 khi x = y = z = 4.
\(P=\dfrac{4x}{2.2.\sqrt{y+z-4}}+\dfrac{4y}{2.2.\sqrt{x+z-4}}+\dfrac{4z}{2.2.\sqrt{x+y-4}}\)
\(P\ge4\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)\ge4.\dfrac{3}{2}=6\)
Dấu "=" xảy ra khi \(x=y=z=4\)
Ta có \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Leftrightarrow\left(x-2\sqrt{xy}+y\right)+\left(y-2\sqrt{yz}+z\right)+\left(z-2\sqrt{zx}+x\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\ge0\) (luôn đúng)
Vậy \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)
Theo BĐT Cauchy-Schwarz dạng Engel
\(A=\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{z+x}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{1}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+y}=\dfrac{y}{y+z}=\dfrac{z}{z+x}\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:
\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)
\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{x+z}\ge\dfrac{x^2}{x+y+z}+\dfrac{y^2}{x+y+z}+\dfrac{z^2}{x+y+z}=\dfrac{x^2+y^2+z^2}{x+y+z}=\dfrac{\left(x+y+z\right)^2-2\left(\sqrt{xy}+\sqrt{zx}+\sqrt{yz}\right)}{x+y+z}\ge\dfrac{1-2.1}{1}=-1\)Áp dụng bất đẳng thức cô-si ta có:
\(x+y\ge2\sqrt{xy}\) , \(x+z\ge2\sqrt{xz}\) , \(y+z\ge2\sqrt{yz}\)
Cộng vế với vế suy ra:
\(2\left(x+y+z\right)\ge2\left(\sqrt{xy}+\sqrt{zx}+\sqrt{yz}\right)\\ \Leftrightarrow x+y+z\ge1\)
Vậy
Trà ơi ! Mình xin lỗi bạn nhiều lắm bài đó mình lỡ giải sai, để mình sữa lại cho bạn:
Đầu tiên ta vẫn có:\(x+y+z\ge1\) (chứng minh trên)
Vậy \(\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{z+x}\ge\dfrac{x^2}{x+y+z}+\dfrac{y^2}{x+y+z}+\dfrac{z^2}{x+y+z}=\dfrac{x^2+y^2+z^2}{x+y+z}\ge x^2+y^2+z^2\ge0\)
Áp dụng bđt Cauchy-Schwarz:
\(A=\dfrac{1}{\sqrt{x\left(y+2z\right)}}+\dfrac{1}{\sqrt{y\left(z+2x\right)}}+\dfrac{1}{\sqrt{z\left(x+2y\right)}}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}}\)
\(=\dfrac{9}{\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}}\)
Áp dụng liên tiếp Bunyakovsky và AM-GM:
\(\left(\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left[x\left(y+2z\right)+y\left(z+2x\right)+z\left(x+2y\right)\right]\)
\(=3.3\left(xy+yz+xz\right)\)
Mà \(3\left(xy+yz+xz\right)\le\left(x+y+z\right)^2=3\)
\(3.3\left(xy+yz+xz\right)\le3.3=9\)
\(\Leftrightarrow\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+z\sqrt{\left(x+2y\right)}\le\sqrt{9}=3\)
\(\Leftrightarrow A\ge\dfrac{9}{3}=3."="\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)
\(P^2=\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}+\dfrac{2xy}{\sqrt{yz}}+\dfrac{2yz}{\sqrt{zx}}+\dfrac{2zx}{\sqrt{xy}}\)
\(P^2=\left(\dfrac{x^2}{y}+\dfrac{xy}{\sqrt{yz}}+\dfrac{xy}{\sqrt{yz}}+z\right)+\left(\dfrac{y^2}{z}+\dfrac{yz}{\sqrt{zx}}+\dfrac{yz}{\sqrt{zx}}+x\right)+\left(\dfrac{z^2}{x}+\dfrac{zx}{\sqrt{xy}}+\dfrac{zx}{\sqrt{xy}}+y\right)-\left(x+y+z\right)\)
\(P^2\ge4\sqrt[4]{\dfrac{x^4y^2z}{y^2z}}+4\sqrt[4]{\dfrac{y^4z^2x}{z^2x}}+4\sqrt[4]{\dfrac{z^4x^2y}{x^2y}}-\left(x+y+z\right)=3\left(x+y+z\right)\ge36\)
\(\Rightarrow P\ge6\)
\(P_{min}=6\) khi \(x=y=z=4\)