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Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có
\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)
\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)
Câu a :
\(\left\{{}\begin{matrix}\left(x^2+1\right)\left(y^2+1\right)=10\\\left(x+y\right)\left(xy-1\right)=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2+x^2+y^2=9\\\left(x+y\right)\left(xy-1\right)=3\end{matrix}\right.\)
Đặt \(x+y=S\) ; \(xy=P\) , phương trình trở thành :
\(\left\{{}\begin{matrix}S^2-2P+P^2=9\\S\left(P-1\right)=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{3}{P-1}\right)^2-2P+P^2=9\\S=\dfrac{3}{P-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}P=0\\P=-2\\P=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}S=-3\\S=-1\\S=3\end{matrix}\right.\)
Với \(S=-3\) và \(P=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-3\\xy=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=0\end{matrix}\right.\end{matrix}\right.\)
Với \(S=-1\) và \(P=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\xy=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\end{matrix}\right.\)
Với \(S=3\) và \(P=2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\end{matrix}\right.\)
Vậy phương trình có các cặp nghiệm là : \(\left(x;y\right)=\left(0;-3\right)\) ; \(\left(x;y\right)=\left(-3;0\right)\) ; \(\left(x;y\right)=\left(1;-2\right)\) ; \(\left(x;y\right)=\left(-2;1\right)\) ; \(\left(x;y\right)=\left(2;1\right)\) ; \(\left(x;y\right)=\left(1;2\right)\)
Wish you study well !!
Phùng Khánh Linh Ko đúng đâu ! Bạn thay \(x=y=\dfrac{1}{2}\) vào thì ra tới 10 lận . \(\dfrac{1}{\dfrac{1}{2}}+\dfrac{4}{\dfrac{1}{2}}=10\) lận cơ ?
Ta có \(\left(x^2+\dfrac{1}{y^2}\right)\left(y^2+\dfrac{1}{x^2}\right)=x^2y^2+1+1+\dfrac{1}{x^2y^2}=x^2y^2+2+\dfrac{1}{x^2y^2}=\dfrac{x^4y^4+2x^2y^2+1}{x^2y^2}=\dfrac{\left(x^2y^2+1\right)^2}{\left(xy\right)^2}=\left(\dfrac{x^2y^2+1}{xy}\right)^2=\left(xy+\dfrac{1}{xy}\right)^2=\left(xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}\right)^2\)
Áp dụng bđt cosi, ta có \(xy+\dfrac{1}{16xy}\ge2\sqrt{xy.\dfrac{1}{16xy}}=2\sqrt{\dfrac{1}{16}}=2.\dfrac{1}{4}=\dfrac{1}{2}\)
\(2\sqrt{xy}\le\left(x+y\right)^2\Leftrightarrow\sqrt{xy}\le\dfrac{\left(x+y\right)^2}{2}=\dfrac{1}{2}\Leftrightarrow xy\le\dfrac{1}{4}\Leftrightarrow\dfrac{15}{16xy}\ge\dfrac{15}{4}\)
Vậy \(xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}\ge\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{17}{4}\Leftrightarrow\left(xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}\right)^2\ge\dfrac{289}{16}\)
Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}x+y=1\\xy=\dfrac{1}{16xy}\\x=y\end{matrix}\right.\)\(\Leftrightarrow\)\(x=y=0,5\)
Vậy GTNN của \(\left(x^2+\dfrac{1}{y^2}\right)\left(y^2+\dfrac{1}{x^2}\right)\)=\(\dfrac{289}{16}\) và xảy ra khi x=y=0,5
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$\text{VT}(1^2+1^2+1^2)\geq (1+\frac{x}{y+z}+1+\frac{y}{x+z}+1+\frac{z}{x+y})^2$
$\Leftrightarrow 3\text{VT}\geq (3+\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y})^2$
$ = \left[3+\frac{x^2}{xy+xz}+\frac{y^2}{yz+yx}+\frac{z^2}{zy+zx}\right]^2$
$\geq \left[3+\frac{(x+y+z)^2}{2(xy+yz+xz)}\right]^2$
$\geq \left[3+\frac{3(xy+yz+xz)}{2(xy+yz+xz)}\right]^2=\frac{81}{4}$
$\Rightarrow \text{VT}\geq \frac{27}{4}$
Dấu "=" xảy ra khi $x=y=z>0$
Áp dụng BĐT Bunhiacopxky:
VT(12+12+12)≥(1+��+�+1+��+�+1+��+�)2VT(12+12+12)≥(1+y+zx+1+x+zy+1+x+yz)2
⇔3VT≥(3+��+�+��+�+��+�)2⇔3VT≥(3+y+zx+x+zy+x+yz)2
=[3+�2��+��+�2��+��+�2��+��]2=[3+xy+xzx2+yz+yxy2+zy+zxz2]2
≥[3+(�+�+�)22(��+��+��)]2≥[3+2(xy+yz+xz)(x+y+z)2]2
≥[3+3(��+��+��)2(��+��+��)]2=814≥[3+2(xy+yz+xz)3(xy+yz+xz)]2=481
⇒VT≥274⇒VT≥427
Dấu "=" xảy ra khi �=�=�>0x=y=z>0
Hướng dẫn: đặt \(A=\dfrac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
Khi đó \(F-A=x-y+y-z+z-x=0\Rightarrow F=A\)
\(\Rightarrow2F=F+A=\sum\dfrac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x+y\right)^2\left(x^2+y^2\right)}{4\left(x^2+y^2\right)\left(x+y\right)}\)
\(\Rightarrow2F\ge\dfrac{x+y+z}{2}\Rightarrow F\ge\dfrac{x+y+z}{4}\)
Ta có BĐT:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(\Leftrightarrow6\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)+2016\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow7.\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le2016\)
Xét \(P=\frac{1}{\sqrt{3\left(2x^2+y^2\right)}}+\frac{1}{\sqrt{3\left(2y^2+z^2\right)}}+\frac{1}{\sqrt{3\left(2z^2+x^2\right)}}\)
\(P^2=\left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2x^2+y^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2y^2+z^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2z^2+x^2}}\right)^2\)
Áp dụng BĐT Bunhiacopxki ta có:
\(P^2\le\left(\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2\right)\left(\left(\frac{1}{\sqrt{2x^2+y^2}}\right)^2+\left(\frac{1}{\sqrt{2y^2+z^2}}\right)^2+\left(\frac{1}{\sqrt{2z^2+x^2}}\right)^2\right)\)
\(\Leftrightarrow P^2\le\frac{1}{2x^2+y^2}+\frac{1}{2y^2+z^2}+\frac{1}{2z^2+x^2}\)
Mặt khác ta có:
\(\frac{1}{2x^2+y^2}=\frac{1}{x^2+x^2+y^2}\le\frac{1}{9}\left(\frac{1}{x^2}+\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\frac{1}{2y^2+z^2}\le\frac{1}{9}\left(\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
\(\frac{1}{2z^2+x^2}\le\frac{1}{9}\left(\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}\right)\)
\(\Rightarrow P^2\le\frac{1}{3}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le\frac{1}{3}.2016=672\)
\(\Rightarrow P\le4\sqrt{42}\)
Dấu '=' xảy ra khi \(x=y=z=\sqrt{\frac{1}{672}}\)
Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)