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Ta có \(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge\frac{x+y+z}{2}\)
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z\ge\frac{x+y+z}{2}+x+y+z\)
\(\Rightarrow x\left(\frac{x}{y+z}+1\right)+y\left(\frac{y}{x+z}+1\right)+z\left(\frac{z}{x+y}+1\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Rightarrow x\left(\frac{x+y+z}{y+z}\right)+y\left(\frac{y+x+z}{x+z}\right)+z\left(\frac{z+x+y}{x+y}\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Rightarrow\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge\frac{3}{2}\) (Theo BĐT Nesbitt )
\(\Rightarrow\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge\frac{3}{2}\) (đpcm)
\(BDT\Leftrightarrow\text{∑}\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)\ge\frac{21}{2}\)
Mà \(\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge2\)(dùng AM-GM giải quyết chỗ này)
Vậy ta cần chứng minh \(\frac{y^2}{z^2}+\frac{z^2}{y^2}+\frac{z^2}{x^2}+\frac{x^2}{z^2}\ge\frac{17}{2}\)
\(\Leftrightarrow\frac{y^2}{z^2}+\frac{x^2}{z^2}\ge\frac{1}{2}\left(\frac{x}{z}+\frac{y}{z}\right)^2\)
\(\Leftrightarrow\frac{z^2}{y^2}+\frac{z^2}{x^2}\ge\frac{1}{2}\left(\frac{4z}{x+y}\right)^2\)
Đặt \(a=\frac{z}{x+y}\ge1\),ta chứng minh \(\frac{1}{2a^2}+8a^2\ge\frac{17}{2}\)
Dễ thấy BĐT này đúng.Vậy ta có đpcm
\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)
\(\Rightarrow xyz\le1\)
\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)
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
Gọi \(\overrightarrow{1a}=\left(x;\frac{1}{x}\right);\overrightarrow{b}=\left(y;\frac{1}{y}\right);\overrightarrow{c}=\left(z;\frac{1}{z}\right)\)
Ta có:
\(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}=\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|\)
\(\ge\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|=\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)\(\ge\sqrt{1^2+\frac{9^2}{\left(x+y+z\right)^2}}\)
\(=\sqrt{1+81}=\sqrt{82}\)
Áp dụng BDT MInkopki
VT\(\ge\)\(\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)\(\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}=\sqrt{82}\)
BDT minkopki
\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}+\sqrt{e^2+f^2}\ge\sqrt{\left(a+c+e\right)^2+\left(b+d+f\right)^2}\)
\(\Sigma\frac{x^3}{y^2}=\Sigma\frac{x}{y^2}\left(x-y\right)^2+\frac{\Sigma z\left(x^3-yz^2\right)^2}{xyz\left(x+y+z\right)}+\Sigma\frac{x^2}{y}\ge\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\)
Áp dụng BĐT Cauchy cho hai số dương :
\(\frac{x^3}{y^2}+x\ge2\sqrt{\frac{x^3}{y^2}\cdot x}=2\frac{x^2}{y}\)
\(\frac{y^3}{z^2}+y\ge2\frac{y^2}{z}\)
\(\frac{z^3}{x^2}+z\ge2\frac{z^2}{x}\)
\(\Rightarrow\frac{x^3}{y^2}+\frac{y^3}{z^2}+\frac{z^3}{x^2}\ge2\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)-\left(x+y+z\right)\)
Mà \(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\ge\frac{\left(x+y+z\right)^2}{x+y+z}=x+y+z\)
\(\Rightarrow\frac{x^3}{y^2}+\frac{y^3}{z^2}+\frac{z^3}{y^2}\ge\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\)
Trước hết ta chứng minh bổ đề sau đây: \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{9\left(a^2+b^2+c^2\right)}{\left(a+b+c\right)^2}\). Đặt P = VT - VP.
(đây là phân tích của một người khác, không phải của em)
Do đó \(VT=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\frac{9\left(x^2+y^2+z^2\right)}{\left(x+y+z\right)^2}=\frac{27}{\sqrt{\left(x+y+z\right)^2.\left(x+y+z\right)^2}}\)
\(\ge\frac{27}{\sqrt{3\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2}}=\frac{9}{x+y+z}\)
Đẳng thức xảy ra khi x = y = z = 1
P/s: Em không chắc lắm!
Theo giả thiết: \(x^2+y^2+z^2=3\Rightarrow2\left(xy+yz+zx\right)=\left(x+y+z\right)^2-3\)
Theo BĐT Bunyakovsky dạng phân thức, ta có:
\(VT=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{x^2}{xy}+\frac{y^2}{yz}+\frac{z^2}{zx}\)\(\ge\frac{\left(x+y+z\right)^2}{xy+yz+zx}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3}\)
Đến đây, ta cần chỉ ra rằng \(\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3}\ge\frac{9}{x+y+z}\)(*)
Ta có: \(xy+yz+zx>0\Leftrightarrow\left(x+y+z\right)^2\ge x^2+y^2+z^2=3\)
\(\Rightarrow x+y+z>\sqrt{3}\)
Đặt \(x+y+z=t>\sqrt{3}\). Khi đó (*) trở thành \(\frac{2t^2}{t^2-3}\ge\frac{9}{t}\Leftrightarrow\frac{\left(t-3\right)^2\left(2t+3\right)}{t\left(t^2-3\right)}\ge0\)(đúng với mọi \(t>\sqrt{3}\))
Đẳng thức xảy ra khi \(t=3\)hay x = y = z = 1