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\(A\le\sqrt{3\left(x+y+y+z+z+x\right)}=\sqrt{6\left(x+y+z\right)}\le\sqrt{6.\sqrt{3\left(x^2+y^2+z^2\right)}}=\sqrt{6\sqrt{3}}\)
\(A_{max}=\sqrt{6\sqrt{3}}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)
Do \(x^2+y^2+z^2=1\Rightarrow0\le x;y;z\le1\)
\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\\z^2\le z\end{matrix}\right.\) \(\Rightarrow x+y+z\ge x^2+y^2+z^2=1\)
\(A^2=2\left(x+y+z\right)+2\sqrt{\left(x+y\right)\left(x+z\right)}+2\sqrt{\left(x+y\right)\left(y+z\right)}+2\sqrt{\left(y+z\right)\left(z+x\right)}\)
\(A^2=2\left(x+y+z\right)+2\sqrt{x^2+xy+yz+zx}+2\sqrt{y^2+xy+yz+zx}+2\sqrt{z^2+xy+yz+zx}\)
\(A^2\ge2\left(x+y+z\right)+2\sqrt{x^2}+2\sqrt{y^2}+2\sqrt{z^2}=4\left(x+y+z\right)\ge4\)
\(\Rightarrow A\ge2\)
\(A_{min}=2\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị
Với a,b,c dưog thì \(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}>=\dfrac{\left(x+y+z\right)^2}{a+b+c}\)
\(P>=\dfrac{\left(x+y+z\right)^2}{xy+yz+xz+\sqrt{1+x^3}+\sqrt{1+y^3}+\sqrt{1+z^3}}\)
\(\sqrt{1+x^3}=\sqrt{\left(1+x\right)\left(1-x+x^2\right)}< =\dfrac{2+x^2}{2}\)
Dấu = xảy ra khi x=2
=>\(P>=\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x^2+y^2+z^2+6}=\dfrac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2+6}\)
Đặt t=(x+y+z)^2(t>=36)
=>P>=2t/t-6
Xét hàm số \(f\left(t\right)=\dfrac{t}{t+6}\left(t>=36\right)\)
\(f'\left(t\right)=\dfrac{6}{\left(t+6\right)^2}>=0,\forall t>=36\)
=>f(t) đồng biến
=>f(t)>=f(36)=6/7
=>P>=12/7
Dấu = xảy ra khi x=y=z=2
áp dụng bđt cô si ta có:
\(xy\le\frac{x^2+y^2}{2};yz\le\frac{y^2+z^2}{2};zx\le\frac{z^2+x^2}{2}\)
\(\Rightarrow A\ge\sqrt{\frac{x^2+y^2}{2}}+\sqrt{\frac{y^2+z^2}{2}}+\sqrt{\frac{z^2+x^2}{2}}\)
theo bunhia thì \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2;2\left(y^2+z^2\right)\ge\left(y+z\right)^2;2\left(z^2+x^2\right)\ge\left(z+x\right)^2\)
\(\Rightarrow A\ge\sqrt{\frac{\left(x+y\right)^2}{4}}+\sqrt{\frac{\left(y+z\right)^2}{4}}+\sqrt{\frac{\left(z+x\right)^2}{4}}=\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=x+y+z=1\)
Vậy \(Min_A=1\Leftrightarrow x=y=z=\frac{1}{3}\)
\(VT=6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(=6\left(x+y+z\right)^2-2\left(xy+yz+xz\right)+2\frac{9}{2x+y+z+x+2y+z+x+y+2z}\)
\(\ge6\left(x+y+z\right)^2-2\frac{\left(x+y+z\right)^2}{3}+2\frac{9}{4\left(x+y+z\right)}\)
\(=\: 6\cdot\left(\frac{3}{4}\right)^2-2\cdot\frac{\left(\frac{3}{4}\right)^2}{3}+2\cdot\frac{9}{4\cdot\frac{3}{4}}=9\)
Có \(\sqrt{\dfrac{xy}{x+y+2z}}=\dfrac{\sqrt{xy}}{\sqrt{x+y+2z}}\)\(=\dfrac{2\sqrt{xy}}{\sqrt{\left(1+1+2\right)\left(x+y+2z\right)}}\)\(\le\dfrac{2\sqrt{xy}}{\sqrt{x}+\sqrt{y}+2\sqrt{z}}\) (theo bunhia dưới mẫu)\(\le\dfrac{2\sqrt{xy}}{4}\left(\dfrac{1}{\sqrt{x}+\sqrt{z}}+\dfrac{1}{\sqrt{y}+\sqrt{z}}\right)\)
\(\Leftrightarrow\sqrt{\dfrac{xy}{x+y+2z}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}}{\sqrt{y}+\sqrt{z}}\right)\)
Tương tự cũng có:
\(\sqrt{\dfrac{yz}{y+z+2x}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{yz}}{\sqrt{y}+\sqrt{x}}+\dfrac{\sqrt{yz}}{\sqrt{z}+\sqrt{x}}\right)\)
\(\sqrt{\dfrac{zx}{z+x+2y}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{zx}}{\sqrt{z}+\sqrt{y}}+\dfrac{\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)
Cộng vế với vế ta được:
\(VT\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}+\sqrt{yz}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}+\sqrt{zx}}{\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{yz}+\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)
\(\Leftrightarrow VT\le\dfrac{1}{2}\left(\sqrt{y}+\sqrt{x}+\sqrt{z}\right)=\dfrac{1}{2}\)
Dấu = xảy ra khi \(x=y=z=\dfrac{1}{9}\)
