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Vì \(\left\{{}\begin{matrix}x>y\\xy< 0\end{matrix}\right.\)\(\Rightarrow x>0>y\)
Đặt \(y=-z\left(z>0\right)\) thì ta có:
\(P=\left(x+z\right)^2+\left(x+z+\dfrac{1}{x}+\dfrac{1}{z}\right)^2\)
\(\ge\left(x+z\right)^2+\left(x+z+\dfrac{4}{x+z}\right)^2\)
Đặt \(x+z=a\) thì ta có:
\(P\ge a^2+\left(a+\dfrac{4}{a}\right)^2=2a^2+\dfrac{16}{a^2}+8\)
\(\ge8+2\sqrt{2a^2.\dfrac{16}{a^2}}=8+8\sqrt{2}\)
Dấu = xảy ra khi: \(\left\{{}\begin{matrix}x=z\\2a^2=\dfrac{16}{a^2}\end{matrix}\right.\)
\(\Rightarrow x=z=\dfrac{1}{\sqrt[4]{2}}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{\sqrt[4]{2}}\\y=-\dfrac{1}{\sqrt[4]{2}}\end{matrix}\right.\)
câu 1:
\(a^2+1\ge2a\\ b^2+1\ge2b\\ c^2+1\ge2c\\ a^2+b^2\ge2ab\\ b^2+c^2\ge2bc\\ a^2+c^2\ge2ac\\ \Rightarrow3\left(a^2+b^2+c^2\right)+3\ge2\left(a+b+c+ab+bc+ac\right)=2.6=12\\ \Rightarrow a^2+b^2+c^2\ge3\)
Dấu "=" xảy ra khi a=b=c=1
Câu 2)
Có \(P=\dfrac{1}{2\left(x^2+y^2\right)}+\dfrac{4}{xy}+2xy\)
\(P=\dfrac{1}{2\left(x^2+y^2\right)}+\dfrac{1}{4xy}+\dfrac{1}{8xy}+\dfrac{29}{8xy}+2xy\)
\(P=\dfrac{1}{2}\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\left(\dfrac{1}{8xy}+2xy\right)+\dfrac{29}{8xy}\)
Áp dụng bất đẳng thức \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) và bất đẳng thức Cô-si, ta được:
\(P\ge\dfrac{1}{2}.\left(\dfrac{4}{\left(x+y\right)^2}\right)+2\sqrt{\dfrac{1}{8xy}.2xy}+\dfrac{29}{2\left(x+y\right)^2}\)
Mà \(x+y\le1\)
\(\Rightarrow P\ge\dfrac{1}{2}.4+2.\dfrac{1}{2}+\dfrac{29}{2}=\dfrac{35}{2}\)
Vậy GTNN của P = \(\dfrac{35}{2}\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=\dfrac{1}{2}.\)
Chúc bạn học tốt!
\(Q=\dfrac{xyz}{z^3\left(x+y\right)}+\dfrac{xyz}{x^3\left(y+z\right)}+\dfrac{xyz}{y^3\left(x+z\right)}\)
\(=\dfrac{1}{z^3\left(x+y\right)}+\dfrac{1}{y^3\left(x+z\right)}+\dfrac{1}{x^3\left(y+z\right)}\) (vì xyz = 1)
\(=\dfrac{\left(\dfrac{1}{z}\right)^2}{z\left(x+y\right)}+\dfrac{\left(\dfrac{1}{y}\right)^2}{y\left(x+z\right)}+\dfrac{\left(\dfrac{1}{x}\right)^2}{x\left(y+z\right)}\)
Áp dụng BĐT cauchy schwarz với x,y,z > 0 ta có:
\(Q\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(xy+yz+xz\right)}=\dfrac{\left(xy+yz+xz\right)^2}{2\left(xy+yz+xz\right)}=\dfrac{xy+yz+xz}{2}\)Mặt khác theo BĐT cauchy với x;y;z>0 thì
\(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}=3\)
Vậy MinQ = \(\dfrac{3}{2}\Leftrightarrow x=y=z=1\)
có x+y=1 =>\(\left\{{}\begin{matrix}x-1=-y\\y-1=-x\end{matrix}\right.\)khí đó ta có biểu thức tương đương :
\(\dfrac{\left(x^2-1\right)\left(y^2-1\right)}{x^2y^2}=\dfrac{\left(x-1\right)\left(x+1\right)\left(y-1\right)\left(y+1\right)}{x^2y^2}=\dfrac{\left(-y\right)\left(x+1\right)\left(-x\right)\left(y+1\right)}{x^2y^2}=\dfrac{\left(x+1\right)\left(y+1\right)}{xy}=\dfrac{xy+x+y+1}{xy}=1+\dfrac{2}{xy}\)mà 1=x+y và x+y\(\ge\)2\(\sqrt{xy}\)=> (x+y)2 \(\ge\)4xy do đó 1= (x+y)2 \(\ge\)4xy
=> \(\dfrac{1}{4xy}\ge\dfrac{1}{\left(x+y\right)^2}=>\dfrac{1}{xy}\ge\dfrac{4}{\left(x+y\right)^2}=>\dfrac{2}{xy}\ge8\)=> biểu thức đã cho có GTNN là 9 khi x=y=\(\dfrac{1}{2}\)
Lời giải:
Ta có: \(A=\frac{3}{x^2+y^2}+\frac{4}{xy}=3\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{5}{2xy}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{x^2+y^2}+\frac{1}{2xy}\geq \frac{4}{x^2+y^2+2xy}=\frac{4}{(x+y)^2}=4\)
Áp dụng BĐT Am-Gm: \(xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}\Rightarrow \frac{5}{2xy}\geq 10\)
Do đó: \(A\geq 3.4+10\Leftrightarrow A\geq 22\)
Vậy \(A_{\min}=22\Leftrightarrow x=y=\frac{1}{2}\)
\(P=\dfrac{1}{2\left(x^2+y^2\right)}+\dfrac{4}{xy}+2xy\)
\(\Leftrightarrow2P=\dfrac{1}{x^2+y^2}+\dfrac{8}{xy}+4xy\)
\(\Leftrightarrow2P=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}+4xy+\dfrac{29}{4xy}\)
Áp dụng BĐT AM - GM , ta có :
\(\Leftrightarrow\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}+4xy+\dfrac{29}{4xy}\ge\dfrac{2}{\sqrt{\left(x^2+y^2\right)2xy}}+2\sqrt{\dfrac{1}{4xy}.4xy}+\dfrac{29}{4xy}\)
\(\Leftrightarrow2P\ge\)\(\dfrac{2}{\sqrt{\left(x^2+y^2\right)2xy}}+2+\dfrac{29}{4xy}\ge\dfrac{4}{\left(x+y\right)^2}+2+\dfrac{29}{\left(x+y\right)^2}\)
\(\Leftrightarrow2P\ge2+4+29=35\)
\(\Leftrightarrow P\ge\dfrac{35}{2}\)
\(\Rightarrow P_{Min}=\dfrac{35}{2}\Leftrightarrow x=y=\dfrac{1}{2}\)