Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
BĐT tương đương với:
\(x+y+z+xy+yz+zx+1\ge3xyz\)
hay : \(7+z\left(6-z\right)+xy\left(1-3z\right)\ge0\)
Vì \(x\le1;y\le2\)nên \(z\ge3\), tức là \(1-3z< 0;3z-5>0\)
Áp dụng BĐT AM-GM, ta có:
\(xy=\frac{1}{2}.2x.y\le\frac{\left(2x+y\right)^2}{8}\le\frac{\left(1+x+y\right)^2}{8}=\frac{\left(7-z\right)^2}{8}\)
Do đó: \(7+z\left(6-z\right)+xy\left(1-3z\right)\ge7+z\left(6-z\right)+\frac{\left(7-z\right)^2}{8}\left(1-3z\right)\)
\(=\frac{1}{8}\left(z-3\right)\left(7-z\right)\left(3z-5\right)=\frac{1}{8}\left(z-3\right)\left(1+x+y\right)\left(3z-5\right)\ge0\)
Đẳng thức xảy ra khi và chỉ khi x=1,y=2,z=3
\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0=>x^2+y^2\ge2xy\\\left(x+y\right)^2\ge0=>x^2+y^2\ge-2xy\end{matrix}\right.\)
Ta có:
\(\left\{{}\begin{matrix}2\left(x^2+y^2\right)+xy\ge5xy\\2\left(x^2+y^2\right)+xy\ge-3xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1\ge5xy\\1\ge-3xy\end{matrix}\right.\)
\(\Leftrightarrow-\dfrac{1}{3}\le xy\le\dfrac{1}{5}\)
Ta có:
P=\(2\left(x^2+y^2\right)^2-4x^2y^2+2+\left(x^2+y^2+2xy\right)\)
P= \(\dfrac{2\left(1-xy\right)^2}{4}-4\left(xy\right)^2+2+\left(\dfrac{1-xy}{2}+2xy\right)\)
=\(\dfrac{\left(xy\right)^2-2xy+1}{2}-4\left(xy\right)^2+2+\dfrac{3xy}{2}+\dfrac{1}{2}\)
Đặt t = xy => \(-\dfrac{1}{3}\le t\le\dfrac{1}{5}\)
Ta có :
P= \(\dfrac{-7t^2}{2}+\dfrac{t}{2}+3=-\dfrac{7}{2}\left(t-\dfrac{1}{14}\right)^2+\dfrac{169}{56}\)
Ta có: \(-\dfrac{1}{3}-\dfrac{1}{14}\le t-\dfrac{1}{14}\le\dfrac{1}{5}-\dfrac{1}{14}\)
<=>\(-\dfrac{17}{42}\le t-\dfrac{1}{14}\le\dfrac{9}{70}\)
=> 0\(\le\left(t-\dfrac{1}{14}\right)^2\le\left(\dfrac{17}{42}\right)^2\)
\(\dfrac{169}{56}\ge P\ge\dfrac{169}{56}-\dfrac{7}{2}\left(\dfrac{17}{42}\right)^2\)
Max P= \(\dfrac{169}{56}\) => t = 1/14 => \(xy=\dfrac{1}{14}\rightarrow x^2+y^2=\dfrac{13}{14}\) => x,y=...
Min P=\(\dfrac{169}{56}-\dfrac{7}{6}\left(\dfrac{17}{42}\right)^2\) <=> \(t=xy=-\dfrac{1}{3}\)
<=> x=-y=\(\dfrac{1}{\sqrt{3}}\)
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)