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Bunhiacopxki: \(\left(x^2+yz+zx\right)\left(y^2+yz+zx\right)\ge\left(xy+yz+zx\right)^2\)
\(\Rightarrow\frac{xy}{x^2+yz+zx}\le\frac{xy\left(y^2+yz+zx\right)}{\left(xy+yz+zx\right)^2}\)
Thiết lập tương tự và cộng lại:
\(\Rightarrow VT\le\frac{xy\left(y^2+yz+zx\right)+yz\left(z^2+xy+zx\right)+zx\left(x^2+yz+xy\right)}{\left(xy+yz+zx\right)^2}\)
\(VT\le\frac{xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz}{\left(xy+yz+zx\right)^2}\)
Ta chỉ cần chứng minh: \(\frac{xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz}{\left(xy+yz+zx\right)^2}\le\frac{x^2+y^2+z^2}{xy+yz+zx}\)
\(\Leftrightarrow xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz\le\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)\)
\(\Leftrightarrow x^2yz+xy^2z+xyz^2\le x^3y+y^3z+z^3x\)
\(\Leftrightarrow\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}\ge x+y+z\) (đúng theo Cauchy-Schwarz)
Dấu "=" xảy ra khi \(x=y=z\)
BĐT của bạn bị ngược dấu, mà có vẻ các mẫu số cũng ko đúng (để ý mẫu số thứ 2 và thứ 3 đều có chung xy+xz ko hợp lý)
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 !!!
\(\frac{1}{x^2+yz}\le\frac{1}{2\sqrt{x^2.yz}}=\frac{1}{2\sqrt{xy.xz}}\le\frac{1}{4}\left(\frac{1}{xy}+\frac{1}{xz}\right)\)
Tương tự: \(\frac{1}{y^2+zx}\le\frac{1}{4}\left(\frac{1}{xy}+\frac{1}{yz}\right)\) ; \(\frac{1}{z^2+xy}\le\frac{1}{4}\left(\frac{1}{xz}+\frac{1}{yz}\right)\)
Cộng vế với vế ta sẽ có đpcm
\(VT=\frac{\left(yz\right)^2}{x^2yz\left(y+z\right)}+\frac{\left(xz\right)^2}{zxy^2\left(x+z\right)}+\frac{\left(xy\right)^2}{xyz^2\left(x+y\right)}\)
\(VT=\frac{2\left(yz\right)^2}{xy+zx}+\frac{2\left(xz\right)^2}{xy+yz}+\frac{2\left(xy\right)^2}{xz+yz}\ge\frac{2\left(yz+xz+xy\right)^2}{2\left(xy+yz+zx\right)}=xy+yz+zx\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{\sqrt[3]{2}}\)
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
http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%AAn-b%C3%ACnh-thu%E1%BA%ADn-2016-2017/
bài của tui mà -_-