Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\text{VT}=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x^2}{\sqrt[3]{x^3yz}}+\frac{y^2}{\sqrt[3]{y^3xz}}+\frac{z^2}{\sqrt[3]{z^3xy}}\)
\(\geq \frac{(x+y+z)^2}{\sqrt[3]{x^3yz}+\sqrt[3]{y^3xz}+\sqrt[3]{z^3xy}}\) (1)
Áp dụng BĐT Am-Gm:
\(\sqrt[3]{x^3yz}\leq \frac{x^2+xyz+1}{3}; \sqrt[3]{y^3xz}\leq \frac{y^2+xyz+1}{3}; \sqrt[3]{z^3xy}\leq \frac{z^2+xyz+1}{3}\)
\(\Rightarrow \sqrt[3]{x^3yz}+\sqrt[3]{y^3xz}+\sqrt[3]{z^3xy}\leq \frac{x^2+y^2+z^2+3xyz+3}{3}=2+xyz\)
Theo BĐT AM-GM:
\(x^2+y^2+z^2\geq 3\sqrt[3]{x^2y^2z^2}\Leftrightarrow 3\sqrt[3]{x^2y^2z^2}\leq 3\Leftrightarrow xyz\leq 1\)
Do đó: \(\sqrt[3]{x^3yz}+\sqrt[3]{y^3xz}+\sqrt[3]{z^3xy}\leq 3\) (2)
Từ (1),(2) và sử dụng hệ quả \(x^2+y^2+z^2\geq xy+yz+xz\) :
\(\Rightarrow \text{VT}\geq \frac{(x+y+z)^2}{3}=\frac{x^2+y^2+z^2+2(xy+yz+xz)}{3}\geq \frac{3(xy+yz+xz)}{3}=xy+yz+xz\)
Ta có đpcm
Dấu bằng xảy ra khi \(x=y=z=1\)
Áp dụng BĐT AM-GM ta có:
\(VT\ge\dfrac{x}{\dfrac{y+z+1}{3}}+\dfrac{y}{\dfrac{x+z+1}{3}}+\dfrac{z}{\dfrac{x+y+1}{3}}\)
Cần chứng minh \(\dfrac{9x}{y+z+1}+\dfrac{9y}{x+z+1}+\dfrac{9z}{x+y+1}\ge3\left(xy+yz+xz\right)\)
Cauchy-Schwarz: \(VT=\dfrac{9x^2}{xy+xz+x}+\dfrac{9y^2}{xy+yz+y}+\dfrac{9z^2}{xz+yz+z}\)
\(\ge\dfrac{9\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\left(x+y+z\right)^2\)
BĐT cuối đúng vì dễ thấy: \(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)
\(\Rightarrow xyz\le1\)
\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)
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
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 = \(\frac{3}{2}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{yz}+\sqrt{xz}+\sqrt{xy}}\)
Xét \(\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{yz}+\sqrt{xz}+\sqrt{xy}}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}\sqrt{yz}\le\frac{y+z}{2}\\\sqrt{xz}\le\frac{x+z}{2}\\\sqrt{xy}\le\frac{x+y}{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\le\frac{y+z}{2}+\frac{x+z}{2}+\frac{x+y}{2}\)
\(\Rightarrow\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\le\frac{2\left(x+y+z\right)}{2}\)
\(\Rightarrow\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\le x+y+z\)
\(\Rightarrow x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le2\left(x+y+z\right)\)
\(\Rightarrow\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{xz}+\sqrt{yz}}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)
Ta có: \(x+y+z\ge3\)
\(\Rightarrow\frac{x+y+z}{2}\ge\frac{3}{2}\)
\(\Rightarrow\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{xz}+\sqrt{yz}}\ge\frac{3}{2}\)
Vì \(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{yz}+\sqrt{xz}+\sqrt{xy}}\)
\(\Rightarrow\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{3}{2}\) ( đpcm )
\(VT=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\)
\(\ge\frac{3x}{y+z+1}+\frac{3y}{x+z+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+xz+x}+\frac{3y^2}{xy+yz+y}+\frac{3z^2}{xz+yz+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x^2+y^2+z^2}\)
\(\ge\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\ge xy+yz+xz=VP\)
Dấu "=" <=> x=y=z=1
Theo giả thiết xy + yz + zx = 1 nên ta có: \(VT=\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2}=\frac{1}{xy+yz+zx+x^2}+\frac{1}{xy+yz+zx+y^2}+\frac{1}{xy+yz+zx+z^2}=\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+x\right)\left(y+z\right)}+\frac{1}{\left(z+x\right)\left(z+y\right)}=\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)Theo bất đẳng thức Cauchy-Schwarz: \(\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)^2\le\left(x+y+z\right)\left(\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\right)=\left(x+y+z\right)\left(\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(y+z\right)\left(y+x\right)}+\frac{z}{\left(z+x\right)\left(z+y\right)}\right)=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)\(\Rightarrow\frac{2}{3}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)^3\le\frac{4\left(x+y+z\right)}{3\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)\)Ta cần chứng minh: \(\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge\frac{4\left(x+y+z\right)}{3\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)\)
hay \(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\le\frac{3}{2}\)
Bất đẳng thức cuối đúng theo AM - GM do: \(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{y+z}.\frac{y}{x+y}}+\sqrt{\frac{z}{z+x}.\frac{z}{z+y}}\le\frac{\left(\frac{x}{x+y}+\frac{x}{x+z}\right)+\left(\frac{y}{y+z}+\frac{y}{x+y}\right)+\left(\frac{z}{z+x}+\frac{z}{z+y}\right)}{2}=\frac{3}{2}\)Đẳng thức xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
Dễ dàng chứng minh được:
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) với \(a,b,c>0\)(1)
Dấu bằng xảy ra \(\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Theo đề bài, vì x, y, z > 0 nên áp dụng (1), ta có:
\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{zx}}+\frac{z^2}{z+\sqrt{xy}}\ge\)\(\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\)(2)
Vì x y, z > 0 nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:
\(x+y\ge2\sqrt{xy}\)(3)
Chứng mih tương tự, ta được;
\(y+z\ge2\sqrt{yz}\)(4);
\(z+x\ge2\sqrt{zx}\)(5)
Từ (3), (4), (5), ta được:
\(2\left(x+y+z\right)\ge2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
\(\Leftrightarrow x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Leftrightarrow2\left(x+y+z\right)\ge x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Leftrightarrow\frac{1}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\)\(\frac{1}{2\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\frac{x+y+z}{2}\)
Mà theo đề bài, \(x+y+z\ge3\) nên:
\(\frac{x+y+z}{2}\ge\frac{3}{2}\)
Suy ra \(\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\frac{3}{2}\left(6\right)\)
Từ (2) và (6), ta được:
\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{zx}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{3}{2}\)(điều phải chứng minh)
Dấu bằng xảy ra
\(\Leftrightarrow\hept{\begin{cases}x=y=z\\x+y+z=3\end{cases}\Leftrightarrow x=y=z=1}\)
Vậy nếu x, y, z > 0 và \(x+y+z\ge3\)thì \(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{zx}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{3}{2}\)
B1:x^2+2016=xy+yz+xz+x^2=...
tuong tu
y^2+2016=... ; z^2+2016=....
B2:bdt am-gm
Áp dụng BĐT Cauchy - Schwarz ta có :
\(VT=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x^2}{\sqrt[3]{x^3yz}}+\frac{y^2}{\sqrt[3]{y^3xz}}+\frac{z^2}{\sqrt[3]{z^3xy}}\)
\(\ge\frac{\left(x+y+z\right)^2}{\sqrt[3]{x^3yz}+\sqrt[3]{y^3xz}+\sqrt[3]{z^3xy}}\left(1\right)\)
Áp dụng BĐT : AM - GM :
\(\sqrt[3]{x^3yz}\le\frac{x^2+xyz+1}{3};\sqrt[3]{y^3xz}\le\frac{y^2+xyz+1}{3};\sqrt[3]{z^3xy}\le\frac{z^2+xyz+1}{3}\)
\(\Rightarrow\sqrt[3]{x^3yz}+\sqrt[3]{y^3xz}+\sqrt[3]{z^3xy}\le\frac{x^2+y^2+z^2+3xyz+3}{3}=2+xyz\)
Theo BĐT AM - GM :
\(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\Leftrightarrow3\sqrt[3]{x^2y^2z^2}\le3\Leftrightarrow xyz\le1\)
Do đó : \(\sqrt[3]{x^3yz}+\sqrt[3]{y^3xz}+\sqrt[3]{z^3xy}\le3\left(2\right)\)
Tư (1) , (2) và sử dụng hệ quả :
\(x^2+y^2+z^2\ge xy+yz+zx:\)
\(\Rightarrow VT\ge\frac{\left(x+y+z\right)^2}{3}=\frac{x^2+y^2+z^2+2\left(xy+yz+xz\right)}{3}\ge\frac{3\left(xy+yz+xz\right)}{3}\)\(=xy+yz+xz\)
Ta có đpcm
Dấu " = " xảy ra khi \(x=y=z=1\)
Chúc bạn học tốt !!!