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Ta có: \((x + y)xy = x^2 + y^2 - xy \quad (1)\)
\(M = \frac{x^3 + y^3}{x^3 y^3} = \frac{(x + y)(x^2 - xy + y^2)}{(xy)^3}\)
(1): \(\left(x+y\right)\cdot xy=x^2+y^2-xy\)
Do đó: \(M=\frac{xy\left(x+y\right)\left(x+y\right)}{\left(xy\right)^3}=\frac{\left(x+y\right)^2}{\left(xy\right)^2}\)
Đặt \(t = \frac{x + y}{xy} = \frac{1}{x} + \frac{1}{y}\)
=>\(M=t^2\)
(1): \((x + y)xy = x^2 + y^2 - xy\)
=>\((x+y)xy=(x+y)^2-3xy\)
=>\(\frac{x + y}{xy} = \left(\frac{x + y}{xy}\right)^2 - \frac{3}{xy}\)
=>\(t=t^2-\frac{3}{xy}\iff\frac{3}{xy}=t^2-t\iff xy=\frac{3}{t^2 - t}\)
\(\left(x+y\right)^2\ge4xy\)
=>\(\left(t\cdot xy\right)^2\ge4xy\)
=>\(t^2\cdot\left(xy\right)^2-4xy\ge0\)
=>xy\(\left(t\cdot xy-4\right)\) >=0
=>\(\frac{3}{t^2 - t} \left(t^2 \cdot \frac{3}{t^2 - t} - 4\right) \ge 0\)
=>\(\frac{3}{t^2 - t}\left(\frac{3t^2 - 4(t^2 - t)}{t^2 - t}\right)\ge0\)
=>\(\frac{3(4t - t^2)}{(t^2 - t)^2}\ge0\)
=>\(4t-t^2\ge0\)
=>\(t^2-4t\le0\)
=>t(t-4)<=0
=>0<t<=4
=>t∈(0;4]
\(M = t^2 \le 4^2 = 16\)
Dấu '=' xảy ra khi t=4
=>\(xy=\frac{3}{4^2 - 4}=\frac{3}{12}=\frac{1}{4};x+y=t\cdot xy=1\)
=>x,y là các nghiệm của phương trình:
\(A^2-A+\frac14=0\)
=>\(\left(A-\frac12\right)^2=0\)
=>A-1/2=0
=>A=1/2
=>x=y=1/2
\(Ta \) \(có : x^2 +y^2 +xy = 1\)
\(\Leftrightarrow\)\(xy = 1 - x^2 - y^2\)
\(Thay \) \(xy = 1 - x^2 - y^2 \) \(vào \) \(P , ta \) \(được :\)
\(P = 1 - x^2 -y^2\)
\(P = 1 - ( x^2 +y^2 )\)
\(P = - ( x^2 +y^2 )+ 1\)\(\le\)\(1\)
\(Dấu "=" xảy \) \(ra\) \(\Leftrightarrow\)\(x^2+y^2 =0\)
\(\Leftrightarrow\)\(x = 0 \) \(và\) \(y = 0\)
\(Max \) \(P = 1 \)\(\Leftrightarrow\)\(x = 0 ; y = 0\)
\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0=>x^2+y^2\ge2xy\\\left(x+y\right)^2\ge0=>x^2+y^2\ge-2xy\end{matrix}\right.\)
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=\(\dfrac{1}{\sqrt{3}}\)
\(B=\dfrac{1}{x^3+y^3}+\dfrac{1}{xy\left(x+y\right)}=\dfrac{1}{x^3+y^3}+\dfrac{3}{3xy\left(x+y\right)}\)
\(B\ge\dfrac{\left(1+\sqrt{3}\right)^2}{x^3+y^3+3xy\left(x+y\right)}=\dfrac{4+2\sqrt{3}}{\left(x+y\right)^3}=4+2\sqrt{3}\)
\(B_{min}=4+2\sqrt{3}\) khi \(\left(x;y\right)=\left(\dfrac{3+\sqrt{3}-\sqrt[4]{12}}{6+2\sqrt{3}};\dfrac{3+\sqrt{3}+\sqrt[4]{12}}{6+2\sqrt{3}}\right)\) và hoán vị
Lời giải:
Áp dụng BĐT Cauchy-Shwarz:
$B=\frac{1}{x^3+y^3}+\frac{1}{xy}=\frac{1}{(x+y)^3-3xy(x+y)}+\frac{1}{xy}$
$=\frac{1}{1-3xy}+\frac{1}{xy}=\frac{1}{1-3xy}+\frac{3}{3xy}$
$\geq \frac{(1+\sqrt{3})^2}{1-3xy+3xy}=(1+\sqrt{3})^2$
Vậy $B_{\min}=(1+\sqrt{3})^2$
Dấu "=" xảy ra khi $xy=\frac{1}{2}-\frac{1}{2\sqrt{3}}$
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 \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)