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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 \(\Leftrightarrow x=y=\dfrac{1}{2}\)
Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)
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Câu hỏi của Con Heo - Toán lớp 8 - Học trực tuyến OLM
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 $x=y=z>0$
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
C1:
\(x,y>0\)
\(M=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2=x^2+2+\dfrac{1}{x^2}+y^2+2+\dfrac{1}{y^2}=\left(x^2+\dfrac{1}{16x^2}\right)+\left(y^2+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+4\)Theo BĐT AM-GM (Caushy) ta có:
\(M=\left(x^2+\dfrac{1}{16x^2}\right)+\left(y^2+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+4\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}.2\sqrt{\dfrac{1}{x^2}.\dfrac{1}{y^2}}+4=\dfrac{1}{2}+\dfrac{1}{2}+4+\dfrac{15}{4}.\dfrac{1}{xy}\ge5+\dfrac{15}{4}.\dfrac{1}{\left(\dfrac{x+y}{2}\right)^2}\ge5+\dfrac{15}{4}.\dfrac{1}{\left(\dfrac{1}{2}\right)^2}=20\)Đẳng thức xảy ra \(\left\{{}\begin{matrix}x^2=\dfrac{1}{16}x^2\\y^2=\dfrac{1}{16}y^2\\x+y=1\\x,y>0\end{matrix}\right.\Leftrightarrow x=y=\dfrac{1}{2}\)
Vậy \(MinM=20\)
Áp dụng bất đẳng thức AM - GM:
\(P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\).
Áp dụng bất đẳng thức AM - GM ta có:
\(xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}\).
Tương tự: \(yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}\).
Do đó \(\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}\)
\(\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}\).
Mà \(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}\)
Nên \(\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}\)
\(\Rightarrow P\ge\dfrac{15}{2}\).
Vậy...
Áp dụng bất đẳng thức AM - GM:
P≥33√(xy+1)(yz+1)(zx+1)xyz.
Áp dụng bất đẳng thức AM - GM ta có:
xy+1=xy+14+14+14+14≥55√xy44.
Tương tự: yz+1≥55√yz44;zx+1≥55√zx44.
Do đó (xy+1)(yz+1)(zx+1)≥1255√(xyz)2412
⇒(xy+1)(yz+1)(zx+1)xyz≥1255√1412(xyz)3.
Mà xyz≤(x+y+z)327=18
Nên (xy+1)(yz+1)(zx+1)xyz≥1255√83412=1255√1215=1258
⇒P≥152.
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 \(B\ge\dfrac{\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2}{2}\) \(=\dfrac{\left(1+\dfrac{1}{xy}\right)^2}{2}\)
Lại có \(xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\)
\(\Rightarrow B\ge\dfrac{\left(1+4\right)^2}{2}=\dfrac{25}{2}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
Vậy GTNN của B là \(\dfrac{25}{2}\) khi \(x=y=\dfrac{1}{2}\)