Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{27}{3\sqrt{3x-2}+6}+\frac{8+4x-x^2}{x\sqrt{6-x}+4}\ge\frac{3}{2}+\frac{2x-14}{3\sqrt{6-x}+2}>0\)
Nên phần còn lại vô nghiệm
Lời giải:
Đặt biểu thức vế trái là $A$
Áp dụng BĐT Bunhiacopxky:
\(A[x(yz+zt+ty)+y(xz+zt+xt)+z(xt+yt+xy)+t(xy+yz+xz)]\geq \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)^2\)
Vì $xyzt=1$ nên:
\(x(yz+zt+ty)+y(xz+zt+xt)+z(xt+yt+xy)+t(xy+yz+xz)=\frac{1}{t}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{x}+\frac{1}{z}+\frac{1}{y}+\frac{1}{x}+\frac{1}{t}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}=3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)\)
Do đó:
$A. 3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)\geq \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)^2$
$\Rightarrow A\geq \frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}}{3}$
Áp dụng BĐT AM-GM: \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\geq 4\sqrt[4]{\frac{1}{xyzt}}=4$
Vậy $A\geq \frac{4}{3}$ (đpcm)
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 \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)
Lời giải
\(\text{HPT}\Leftrightarrow \left\{\begin{matrix} \frac{xy+yz+xz}{y+z}=\frac{1}{2}\\ \frac{xy+yz+xz}{z+x}=\frac{1}{3}\\ \frac{xy+yz+xz}{x+y}=\frac{1}{4}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{x+z}{y+z}=\frac{3}{2}\\ \frac{x+y}{x+z}=\frac{4}{3}\\ \frac{y+z}{x+y}=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} 2x-3y-z=0\\ -x+3y-4z=0\\ -x+y+2z=0\end{matrix}\right.\Rightarrow 3x=5y=15z\)
Thay vào phương trình ban đầu: \(5z+\frac{3z.z}{3z+z}=\frac{1}{2}\Leftrightarrow z=\frac{2}{23}\Rightarrow x=\frac{10}{23},y=\frac{6}{23}\)
Thử lại thấy đúng
Vậy nghiệm của HPT là \((x,y,z)=(\frac{10}{23},\frac{6}{23},\frac{2}{23})\)
Cho x, y, z >0 thoả mãn x+y+z=1. Cmr: \(\frac{x}{x+yz}+\frac{y}{y+xz}+\frac{z}{z+xy}\le\frac{9}{4}\)
\(VT=\sum\frac{x}{x\left(x+y+z\right)+yz}=\sum\frac{x}{\left(x+y\right)\left(x+z\right)}=\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(VT=\frac{2\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(VT=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y+z\right)\left(xy+yz+zx\right)-xyz}=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+\frac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)-xyz}\)
\(VT\le\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+\frac{1}{9}3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}-xyz}\)
\(VT\le\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+xyz-xyz}=\frac{9}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
\(VT=\sum\frac{x}{3-yz}\le\sum\frac{2x}{6-\left(y^2+z^2\right)}=\sum\frac{2x}{x^2+x^2+y^2+z^2}\le\sum\frac{x^2+1}{x^2+1+2}\)
\(VT\le\frac{1}{4}\sum\left(\frac{x^2+1}{x^2+1}+\frac{x^2+1}{2}\right)=\frac{1}{4}\left(3+\frac{x^2+y^2+z^2+3}{2}\right)=\frac{3}{2}\)
\(\frac{6}{2xy+2yz+2zx}+\frac{2}{x^2+y^2+z^2}\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=8+4\sqrt{3}>14\)
Dấu "=" không xảy ra
Lời giải:
ĐK: $x,y,z\neq 0$
Ta thấy: \(\frac{xy}{z}.\frac{yz}{x}=y^2>0\) với mọi $y\neq 0$
$\Rightarrow \frac{xy}{z}, \frac{yz}{x}$ cùng dấu
Tương tự: $\frac{yz}{x}, \frac{xz}{y}$ cùng dấu
$\Rightarrow \frac{xy}{z}, \frac{yz}{x}, \frac{xz}{y}$ cùng dấu.
Nếu cùng dấu âm thì hiển nhiên tổng của chúng phải âm (vô lý vì $3>0$)
Do đó $\frac{xy}{z}, \frac{yz}{x}, \frac{xz}{y}>0$
$\Rightarrow \frac{xy}{z}.\frac{yz}{x}.\frac{xz}{y}=xyz>0(1)$
Áp dụng BĐT AM-GM:
\(3=\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\geq 3\sqrt[3]{xyz}\Rightarrow xyz\leq 1(2)\)
Từ $(1);(2)\Rightarrow xyz=1$
$\Rightarrow (x,y,z)=(1,1,1); (-1,-1,1); (-1,1,-1); (1,-1,-1)$