với các số thực dương x,y,z và xyz=1
chứng minh đẳng thức
\(\dfrac{\sqrt{x^3+y^3+1}}{xy}+\dfrac{\sqrt{y^3+z^3+1}}{yz}+\dfrac{\sqrt{z^3+x^3+1}}{zx}\ge3\sqrt{3}\)
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Ta có \(x^3+y^3\ge xy\left(x+y\right)\)
\(\Rightarrow1+x^3+y^3\ge xyz+xy\left(x+y+z\right)=xy\left(x+y+z\right)\)
Tương tự ta có
\(VT\ge\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}+\dfrac{\sqrt{yz\left(x+y+z\right)}}{yz}+\dfrac{\sqrt{zx\left(x+y+z\right)}}{zx}\)
\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)
\(=\sqrt{x+y+z}.\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{xyz}}\)
\(\ge\sqrt{3\sqrt[3]{xyz}}.\dfrac{3\sqrt[6]{xyz}}{1}=3\sqrt{3}\)
\("="\Leftrightarrow x=y=z=1\)
\(P\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}+\dfrac{\sqrt{3\sqrt[3]{y^3z^3}}}{yz}+\dfrac{\sqrt{3\sqrt[3]{z^3x^3}}}{zx}\)
\(P\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.zx}}}=3\sqrt{3}\)
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 $x=y=z=1.$
\(\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}=\dfrac{\sqrt{3xy}}{xy}=\dfrac{\sqrt{3}}{\sqrt{xy}}\)
Tương tự \(\dfrac{\sqrt{1+y^3+z^3}}{yz}\ge\dfrac{\sqrt{3}}{\sqrt{yz}};\dfrac{\sqrt{1+x^3+z^3}}{xz}\ge\dfrac{\sqrt{3}}{\sqrt{xz}}\)
\(\Rightarrow VT\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3}.\dfrac{3}{\sqrt[3]{xyz}}=3\sqrt{3}\)
Dấu "=" xảy ra khi x=y=z=1
\(P=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+3}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{3\sqrt{z}}{\sqrt{zx}+3\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{xyz}\sqrt{z}}{\sqrt{zx}+\sqrt{xyz}\sqrt{z}+\sqrt{xyz}}\)
\(=\dfrac{1}{\sqrt{y}+1+\sqrt{yz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{yz}}{1+\sqrt{yz}+\sqrt{y}}\)
\(=\dfrac{1+\sqrt{y}+\sqrt{yz}}{1+\sqrt{y}+\sqrt{yz}}=1\)
\(\Rightarrow\sqrt{10P-1}=\sqrt{10.1-1}=\sqrt{9}=3\)
\(\dfrac{\sqrt{1+x^3+y^3}}{xy}>=\sqrt{\dfrac{3}{xy}}\)
\(\dfrac{\sqrt{1+y^3+z^3}}{yz}>=\sqrt{\dfrac{3}{yz}}\)
\(\dfrac{\sqrt{1+z^3+x^3}}{xz}>=\sqrt{\dfrac{3}{xz}}\)
=>\(VT>=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)=3\sqrt{3}\)
Ta xét BĐT phụ: \(1+x^3+y^3\ge xy\left(x+y+z\right)\)
\(x^3+y^3\ge xy\left(x+y\right)+xyz-1\)
\(x^3+y^3-xy\left(x+y\right)\ge0\)
\(\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\ge0\)
\(\left(x+y\right)\left(x-y\right)^2\ge0\)( Luôn đúng, vậy BĐT phụ đúng)
\(\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}=\sqrt{x+y+z}.\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3\sqrt[3]{xyz}}.\left(3\sqrt[3]{\dfrac{1}{\sqrt{x^2y^2z^2}}}\right)=3\sqrt{3}\)
GTNN của P là \(3\sqrt{3}\Leftrightarrow x=y=z=1\)
Gọi \(A=\sum\dfrac{x^3}{\sqrt{y^2+3}}\)
Theo Holder: \(A.A.\left(\left(y^2+3\right)+\left(z^2+3\right)+\left(x^2+3\right)\right)\ge\left(x^3+y^3+z^3\right)^3\)
\(\Rightarrow A^2\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+9}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}=\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+xy+yz+zx}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}\)
Ta có đánh giá sau: \(x^3+y^3+z^3\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\dfrac{\left(x+y+z\right)^3}{9}\)
\(\Rightarrow A^2\ge\dfrac{\dfrac{\left(x+y+z\right)^3}{9}}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{x+y+z}{12}\ge\dfrac{\sqrt{3\left(xy+yz+zx\right)}}{12}\ge\dfrac{1}{4}\)
\(\Rightarrow A\ge\dfrac{1}{2}\)
Lời giải:
Ta có:
\(3=xy+yz+xz\leq \frac{(x+y+z)^2}{3}\Rightarrow x+y+z\geq 3\)
Áp dụng BĐT AM-GM:
\(x^3+8=(x+2)(x^2-2x+4)\leq \left(\frac{x+2+x^2-2x+4}{2}\right)^2\)
\(\Rightarrow \sqrt{x^3+8}\leq \frac{x^2-x+6}{2}\Rightarrow \frac{x^2}{\sqrt{x^3+8}}\geq \frac{2x^2}{x^2-x+6}\)
Thực hiện tương tự với các phân thức còn lại và cộng theo vế:
\(\Rightarrow \text{VT}\geq \underbrace{2\left(\frac{x^2}{x^2-x+6}+\frac{y^2}{y^2-y+6}+\frac{z^2}{z^2-z+6}\right)}_{M}\)
Áp dụng BĐT Cauchy-Schwarz:
\(M\geq \frac{2(x+y+z)^2}{x^2-x+6+y^2-y+6+z^2-z+6}=\frac{2(x+y+z)^2}{x^2+y^2+z^2-(x+y+z)+18}\)
\(\Leftrightarrow M\geq \frac{2(x+y+z)^2}{(x+y+z)^2-(x+y+z)+12}\) (do $xy+yz+xz=3$)
Mà :
\(\frac{(x+y+z)^2}{(x+y+z)^2-(x+y+z)+12}-1=\frac{(x+y+z)^2+(x+y+z)-12}{(x+y+z)^2-(x+y+z)+12}=\frac{(x+y+z-3)(x+y+z+4)}{(x+y+z)^2-(x+y+z)+12}\geq 0\) do $x+y+z\geq 0$
Do đó: \(M\geq 1\Rightarrow \text{VT}\geq 1\) (đpcm)
Dấu bằng xảy ra khi \(x=y=z=1\)
\(A=\dfrac{\sqrt{x^3+y^3+1}}{xy}+\dfrac{\sqrt{y^3+z^3+1}}{yz}+\dfrac{\sqrt{z^3+x^3+1}}{zx}\)
\(\dfrac{\sqrt{x^3+y^3+1}}{xy}=\dfrac{\sqrt{x^3+y^3+xyz}}{xy}\ge\dfrac{\sqrt{xy\left(x+y\right)+xyz}}{xy}=\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}\ge\dfrac{\sqrt{xy.3^3\sqrt{xyz}}}{xy}=\dfrac{\sqrt{3xy}}{xy}=\dfrac{\sqrt{3}}{\sqrt{xy}}\)
\(\dfrac{\sqrt{y^3+z^3+1}}{yz}\ge\dfrac{\sqrt{3}}{\sqrt{yz}}\)
\(\dfrac{\sqrt{z^3+x^3+1}}{zx}\ge\dfrac{\sqrt{3}}{\sqrt{zx}}\)
\(\Rightarrow A\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.xz}}}=3\sqrt{3}.\sqrt[3]{\dfrac{1}{xyz}}=3\sqrt{3}\)