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Ta có: \(x^2+1\ge2x\Leftrightarrow\left(x-1\right)^2\ge0\)
Tương tự cũng có 2 BĐT còn lại:
\(y^2+1\ge2y;z^2+1\ge2z\)
Và \(x^2+y^2+z^2\ge xy+yz+xz\)
\(\Leftrightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)
Cộng theo vế các BĐT trên ta có:
\(3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+xz\right)\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge12\Leftrightarrow x^2+y^2+z^2\ge3\)
Khi x=y=z=1
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Áp dụng BĐT Cô-si dạng Engel,ta có :
\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}\)
Mà \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le x+y+z\)
\(\Rightarrow\)\(\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\ge\frac{3}{2}\)
Dấu "=" xảy ra khi x = y = z = \(\frac{3}{2}\)
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\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)
\(\Rightarrow xyz\le1\)
\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)
Ta co:
\(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x\sqrt[3]{x}}{\sqrt[3]{xyz}}+\frac{y\sqrt[3]{y}}{\sqrt[3]{xyz}}+\frac{z\sqrt[3]{z}}{\sqrt[3]{xyz}}\)
\(\ge x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\)
\(\Rightarrow3A\ge3\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\ge\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)
\(\ge\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow A\ge xy+yz+zx\)
Áp dụng BĐT Cauchy - Schwarz, ta có: \(3\left(x^2+y^2+z^2\right)=\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3=x^2+y^2+z^2\)(Do \(x^2+y^2+z^2=3\))
Ta có: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}=\frac{x}{\sqrt[3]{yz.1}}+\frac{y}{\sqrt[3]{zx.1}}+\frac{z}{\sqrt[3]{xy.1}}\)
\(\ge\frac{x}{\frac{y+z+1}{3}}+\frac{y}{\frac{z+x+1}{3}}+\frac{z}{\frac{x+y+1}{3}}\)\(=\frac{3x}{y+z+1}+\frac{3y}{z+x+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+zx+x}+\frac{3y^2}{yz+xy+y}+\frac{3z^2}{zx+yz+z}\)\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+\left(x+y+z\right)}\)(Theo BĐT Cauchy - Schwarz dạng Engle)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\)
\(\ge xy+yz+zx\)
Đẳng thức xảy ra khi x = y = z = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{\sqrt{1+x^3+y^3}}{xy}>=\sqrt{\dfrac{3}{xy}}\)
\(\dfrac{\sqrt{1+y^3+z^3}}{yz}>=\sqrt{\dfrac{3}{yz}}\)
\(\dfrac{\sqrt{1+z^3+x^3}}{xz}>=\sqrt{\dfrac{3}{xz}}\)
=>\(VT>=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)=3\sqrt{3}\)
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\(M=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{y^3z^3+x^3z^3+x^3y^3}{x^2y^2z^2}=\frac{\left(yz+xz\right)^3+x^3y^3-3xy^2z^3-3x^2yz^3}{x^2y^2z^2}\)
\(=\frac{\left(yz+xz+xy\right)\left[\left(yz+xz\right)^2+xy\left(yz+xz\right)+x^2y^2\right]-3xyz^2\left(xz+yz\right)}{x^2y^2z^2}\)
\(=\frac{0.\left[\left(yz+xz\right)^2+xy\left(yz+xz\right)+x^2y^2\right]-3xyz^2\left(xz+yz\right)}{x^2y^2z^2}\)
\(=\frac{-3xyz^2\left(xz+yz\right)}{x^2y^2z^2}=\frac{-3\left(xz+yz\right)}{xy}=\frac{-3.\left(-xy\right)}{xy}=3\)
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\(\frac{x^4}{y+3z}+\frac{y+3z}{16}+\frac{1}{4}+\frac{1}{4}\ge4\sqrt[4]{\frac{x^4}{y+3z}.\frac{y+3z}{16}.\frac{1}{4}.\frac{1}{4}}=x\)
\(\Rightarrow\frac{x^4}{y+3z}\ge x-\frac{y+3z}{16}-\frac{1}{2}\)
Tương tự cho 2 BĐT còn lại :
\(\frac{y^4}{z+3x}\ge y-\frac{z+3x}{16}-\frac{1}{2};\frac{z^4}{z+3y}\ge z-\frac{x+3y}{16}-\frac{1}{2}\)
Công theo vế 3 BĐT trên ta được :
\(VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{2}\ge\frac{3}{4}.3-\frac{3}{2}=\frac{3}{4}\)
Đẳng thức xảy ra khi \(x=y=z=1\)
Chúc bạn học tốt !!!
Cách 2:
\(VT\ge\frac{\left(x^2+y^2+z^2\right)^2}{4\left(x+y+z\right)}\ge\frac{\frac{\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2}{3}}{4\left(x+y+z\right)}\ge\frac{\left(xy+yz+zx\right)\left(x+y+z\right)}{12}\)
\(\ge\frac{\left(xy+yz+zx\right)\sqrt{3\left(xy+yz+zx\right)}}{12}\ge\frac{3}{4}\)
Đẳng thức xảy ra khi \(x=y=z=1\)
ĐẶT \(X^2+Y^2+Z^2=A\)
\(3A+12=.......\)