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\(P\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}+\dfrac{\sqrt{3\sqrt[3]{y^3z^3}}}{yz}+\dfrac{\sqrt{3\sqrt[3]{z^3x^3}}}{zx}\)
\(P\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.zx}}}=3\sqrt{3}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Ta có bất đẳng thức sau \(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0.\)
Do đó:
\(P=\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xyz+xy\left(x+y\right)}}{xy}\)
\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\dfrac{1}{\sqrt{xy}}\cdot\dfrac{1}{\sqrt{yz}}\cdot\dfrac{1}{\sqrt{zx}}}=3\sqrt{3}\)
Đẳng thức xảy ra khi $x=y=z=1.$
Ta có: \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge9xyz\)
\(VT=\dfrac{x}{1+yz}+\dfrac{y}{1+xz}+\dfrac{z}{1+xy}\)
\(=\dfrac{x^2}{x+xyz}+\dfrac{y^2}{y+xyz}+\dfrac{z^2}{z+xyz}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3xyz}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+\dfrac{\left(x+y+z\right)\left(xy+yz+xz\right)}{3}}\)
\(=\dfrac{3\left(x+y+z\right)}{4}\). Cần chứng minh:
\(\dfrac{3\left(x+y+z\right)}{4}\ge\dfrac{3\sqrt{3}}{4}\Leftrightarrow x+y+z\ge\sqrt{3}\)
BĐT cuối đúng vì \(x+y+z\ge\sqrt{3\left(xy+yz+xz\right)}=\sqrt{3}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)
Ps: nospoiler
Ta có: \(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2zx}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{zx+2yz}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+2zx+yz+2xy+zx+2yz}=\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\)
Mà ta lại có: \(xy+yz+zx\le x^2+y^2+z^2\)
\(\Rightarrow\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1^2}{3.1}=\dfrac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)
Áp dụng BĐT BSC:
\(F=\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\)
\(\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}\right)\)
\(=\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{4}.4=1\)
\(maxF=1\Leftrightarrow x=y=z=\dfrac{3}{4}\)
\(x^2y+x^2y+y^2z\ge3\sqrt[3]{x^4y^4z}=3\sqrt[3]{x^3y^3.xyz}=3\sqrt[3]{x^3y^3}=3xy\)
Tương tự:
\(y^2z+y^2z+z^2x\ge3yz\)
\(z^2x+z^2x+x^2y\ge3zx\)
Cộng vế:
\(3\left(x^2y+y^2z+z^2x\right)\ge3\left(xy+yz+zx\right)\)
\(\Leftrightarrow x^2y+y^2z+z^2x\ge xy+yz+zx\)
Dấu "=" xảy ra khi \(x=y=z=1\)