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Ta có: \(\frac{x^2}{x^4+yz}\le\frac{x^2}{2\sqrt{x^4.yz}}=\frac{x^2}{2x^2\sqrt{yz}}=\frac{1}{2\sqrt{yz}}\)(BĐt cosi) (1)
CMTT: \(\frac{y^2}{y^4+xz}\le\frac{1}{2\sqrt{xz}}\) (2)
\(\frac{z^2}{z^4+xy}\le\frac{1}{2\sqrt{xy}}\)(3)
Từ (1); (2) và (3) =>A = \(\frac{x^2}{x^4+yz}+\frac{y^2}{y^4+xz}+\frac{z^2}{z^4+xy}\le\frac{1}{2}\left(\frac{1}{\sqrt{xz}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xy}}\right)\)
Áp dụng bđt \(ab+bc+ac\le a^2+b^2+c^2\)
cmt đúng: <=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)(luôn đúng)
Khi đó: A \(\le\frac{1}{2}\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}\cdot\frac{xy+yz+xz}{xyz}\le\frac{1}{2}\cdot\frac{x^2+y^2+z^2}{xyz}=\frac{3xyz}{2xyz}=\frac{3}{2}\)
\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)
\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)
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 BĐT Cô-si,ta có :
x4 + yz \(\ge\)\(2\sqrt{x^4yz}=2x^2\sqrt{yz}\); \(y^4+xz\ge2y^2\sqrt{xz}\); \(z^4+xy\ge2z^2\sqrt{xy}\)
\(\Rightarrow\frac{x^2}{x^4+yz}+\frac{y^2}{y^4+xz}+\frac{z^2}{z^4+xy}\le\frac{x^2}{2x^2\sqrt{yz}}+\frac{y^2}{2y^2\sqrt{xz}}+\frac{z^2}{2z^2\sqrt{xy}}=\frac{1}{2\sqrt{yz}}+\frac{1}{2\sqrt{xz}}+\frac{1}{2\sqrt{xy}}\)
CM : x + y + z \(\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)
\(\frac{x^2}{x^4+yz}+\frac{y^2}{y^4+xz}+\frac{z^2}{z^4+xy}\le\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}.\frac{yz+xz+xy}{xyz}=\frac{1}{2}.\frac{3xyz}{xyz}=\frac{3}{2}\)
Áp dụng BĐT Cauchy cho các cặp số dương, ta có: \(\Sigma\frac{x^2}{x^4+yz}\le\Sigma\frac{x^2}{2x^2\sqrt{yz}}=\Sigma\frac{1}{2\sqrt{yz}}\)
\(\le\frac{1}{4}\Sigma\left(\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=\frac{1}{2}.\frac{xy+yz+zx}{xyz}\le\frac{1}{2}.\frac{x^2+y^2+z^2}{xyz}=\frac{1}{2}.\frac{3xyz}{xyz}=\frac{3}{2}\)
Đẳng thức xảy ra khi x = y = z = 1
\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)
\(\Rightarrow xyz\le1\)
\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)
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/
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 !!!
Đoạn cuối của cô Nguyễn Linh Chi em có 1 cách biến đổi tương đương cũng khá ngắn gọn ạ
\(RHS\ge2\cdot\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)
Theo đánh giá của cô Nguyễn Linh Chi thì \(xy+yz+zx\ge x+y+z\ge3\)
Ta cần chứng minh:\(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\ge\frac{1}{2}\)
Thật vậy,BĐT tương đương với:
\(2\left(x+y+z\right)^2\ge x^2+y^2+z^2-x-y-z+18\)
\(\Leftrightarrow\left(x+y+z\right)^2+x+y+z-12\ge0\)
\(\Leftrightarrow\left(x+y+z+4\right)\left(x+y+z-3\right)\ge0\) ( luôn đúng với \(x+y+z\ge3\) )
=> đpcm
Áp dụng: \(AB\le\frac{\left(A+B\right)^2}{4}\)với mọi A, B
Ta có:
\(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\le\frac{\left(x+2+x^2-2x+4\right)^2}{4}\)
=> \(\sqrt{x^3+8}\le\frac{x^2-x+6}{2}\)
=> \(\frac{x^2}{\sqrt{x^3+8}}\ge\frac{2x^2}{x^2-x+6}\)
Tương tự
=> \(\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}\)
\(\ge\frac{2x^2}{x^2-x+6}+\frac{2y^2}{y^2-y+6}+\frac{2z^2}{z^2-z+6}\)
\(=2\left(\frac{x^2}{x^2-x+6}+\frac{y^2}{y^2-y+6}+\frac{z^2}{z^2-z+6}\right)\)
\(\ge2\frac{\left(x+y+z\right)^2}{x^2-x+6+y^2-y+6+z^2-z+6}\)
\(=2\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)(1)
Ta có: \(x+y+z\le xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\) với mọi x, y, z
=> \(\left(x+y+z\right)^2-3\left(x+y+z\right)\ge0\)
=> \(\left(x+y+z\right)\left(x+y+z-3\right)\ge0\)
=> \(x+y+z\ge3\)với mọi x, y, z dương
Và \(x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+zx\right)\le\left(x+y+z\right)^2-2\left(x+y+z\right)\)
Do đó: \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)
\(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3\left(x+y+z\right)+18}\)
Đặt: x + y + z = t ( t\(\ge3\))
Xét hiệu: \(\frac{t^2}{t^2-3t+18}-\frac{1}{2}=\frac{t^2+3t-18}{t^2-3t+18}=\frac{\left(t-3\right)\left(t+6\right)}{\left(t-\frac{3}{2}\right)^2+\frac{63}{4}}\ge0\)với mọi t \(\ge3\)
Do đó: \(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3\left(x+y+z\right)+18}\ge\frac{1}{2}\)(2)
Từ (1); (2)
=> \(\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}\ge2.\frac{1}{2}=1\)
Dấu "=" xảy ra <=> x= y = z = 1
\(\frac{x}{3-yz}+\frac{y}{3-zx}+\frac{z}{3-xy}\le\frac{x}{3-\frac{y^2+z^2}{2}}+\frac{y}{3-\frac{z^2+x^2}{2}}+\frac{z}{3-\frac{x^2+y^2}{2}}\)
\(=\frac{2x}{3+x^2}+\frac{2y}{3+y^2}+\frac{2z}{3+z^2}\le\frac{2x}{4\sqrt[4]{x^2}}+\frac{2y}{4\sqrt[4]{y^2}}+\frac{2z}{4\sqrt[4]{z^2}}\)
\(=\frac{\sqrt{x}}{2}+\frac{\sqrt{y}}{2}+\frac{\sqrt{z}}{2}\le\frac{x^2+3}{8}+\frac{y^2+3}{8}+\frac{z^2+3}{8}\)
\(=\frac{3}{8}+\frac{9}{8}=\frac{3}{2}\)
cách khác: cũng đến chỗ <= sigma 2x/3+x^2
<= 2x/2(x+1) (do x^2+3=x^2+1+2>=2x+2) <= sigma x/x+1 = 3- sigma (1/x+1)
sigma 1/x+1 >= 9/x+y+z+3 dễ rồi