Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng BĐT Cô - si cho 3 bộ số không âm
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)
\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)
\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)
Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)
Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)
\(P=\frac{1}{2}\left(x^2+y^2+z^2\right)+\frac{x^2+y^2+z^2}{xyz}\)
\(P\ge\frac{3}{2}\sqrt[3]{\left(xyz\right)^2}+\frac{3\sqrt[3]{\left(xyz\right)^2}}{xyz}=\frac{3}{2}\sqrt[3]{\left(xyz\right)^2}+\frac{3}{\sqrt[3]{xyz}}\)
\(P\ge\frac{3}{2}\left(\sqrt[3]{\left(xyz\right)^2}+\frac{1}{\sqrt[3]{xyz}}+\frac{1}{\sqrt[3]{xyz}}\right)\ge\frac{9}{2}\) (AM-GM trực tiếp biểu thức trong ngoặc)
Dấu "=" xảy ra khi \(x=y=z=1\)
\(\left(x^3+y^3\right)\left(x+y\right)=xy\left(1-x\right)\left(1-y\right)\Leftrightarrow\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)=\left(1-x\right)\left(1-y\right)\left(1\right)\)
Ta có : \(\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)\ge4xy\)
và \(\left(1-x\right)\left(1-y\right)=1-\left(x+y\right)+xy\le1-2\sqrt{xy}+xy\)
\(\Rightarrow1-2\sqrt{xy}+xy\ge4xy\Leftrightarrow0\) <\(xy\le\frac{1}{9}\)
Dễ chứng minh : \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\le\frac{1}{1+xy};\left(x,y\in\left(0;1\right)\right)\)
\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\le\sqrt{2\left(\frac{1}{1+x^2}+\frac{1}{1+y^2}\right)}\le\sqrt{2\left(\frac{2}{1+xy}\right)}=\frac{2}{\sqrt{1+xy}}\)
\(3xy-\left(x^2+y^2\right)=xy-\left(x-y\right)^2\le xy\)
\(\Rightarrow P\le\frac{2}{\sqrt{1+xy}}+xy=\frac{2}{\sqrt{1+t}}+t\), \(\left(t=xy\right)\), (0<\(t\le\frac{1}{9}\)
Xét hàm số :
\(f\left(t\right)=\frac{2}{\sqrt{t+1}}+t\) , (0<\(t\le\frac{1}{9}\)
Ta có Max \(f\left(t\right)=f\left(\frac{1}{9}\right)=\frac{6\sqrt{10}}{10}+\frac{1}{9}\), \(t\in\left(0;\frac{1}{9}\right)\)
Đặt \(t=\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{xy}{xy}}=2\) \(\Rightarrow t^2=\frac{x^2}{y^2}+\frac{x^2}{y^2}+2\)
\(\Rightarrow A=f\left(t\right)=3\left(t^2-2\right)-8t+10=3t^2-8t+4\)
Xét hàm \(f\left(t\right)\) trên \([2;+\infty)\)
Có \(a=3>0\) ; \(-\frac{b}{2a}=\frac{8}{6}=\frac{4}{3}< 2\)
\(\Rightarrow f\left(t\right)\) đồng biến trên \([2;+\infty)\)
\(\Rightarrow\min\limits_{[2;+\infty)}f\left(t\right)=f\left(2\right)=0\)
Đặt \(\frac{x}{y}=t\)
Ta có: \(A=3\left(t^2+\frac{1}{t^2}\right)-8\left(t+\frac{1}{t}\right)+10\)
Ta sẽ chứng minh \(A\ge0\)
\(3\left(t^2+\frac{1}{t^2}\right)-8\left(t+\frac{1}{t}\right)\ge-10\)
\(\Leftrightarrow3t^2-8t+5+\frac{3}{t^2}-\frac{8}{t}+5\ge0\)
\(\Leftrightarrow\left(3t-5\right)\left(t-1\right)+\left(\frac{3}{t}-5\right)\left(\frac{1}{t}-1\right)\ge0\)
\(\Leftrightarrow\left(3t-5\right)\left(t-1\right)+\left(\frac{5t-3}{t}\right)\left(\frac{t-1}{t}\right)\ge0\)
\(\Leftrightarrow\left(t-1\right)\left(3t-5+\frac{5t-3}{t^2}\right)\ge0\)
\(\Leftrightarrow\frac{\left(t-1\right)^2\left(3t^2-2t+3\right)}{t^2}\ge0\) (đúng)
Đẳng thức xảy ra khi t = 1 hay x = y
Do đó \(A\ge0\) hay Min A = 0 <=> x = y
P/s: Em ko chắc
\(2x^2+2y^2+2z^2=2xy+2yz+2zx\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\) \(\Rightarrow x=y=z\)
\(A=\left(2015-2014\right)\left(2014-2013\right)\left(2013-2012\right)=1\)
\(\Leftrightarrow Q=\frac{\left(x+\frac{y}{2}+\frac{y}{2}\right)^3}{xy^2}\)
Áp dụng BĐT Cô-si cho 3 số dương:
\(x+\frac{y}{2}+\frac{y}{2}\ge3\sqrt[3]{x.\frac{y}{2}.\frac{y}{2}}=3\sqrt[3]{\frac{xy^2}{4}}\)
\(\Rightarrow\left(x+\frac{y}{2}+\frac{y}{2}\right)^3\ge3.\frac{xy^2}{4}\)
\(\Rightarrow Q\ge\frac{3.\frac{xy^2}{4}}{xy^2}=\frac{3}{4}\)
\("="\Leftrightarrow x=\frac{y}{2}\Leftrightarrow y=2x\)