Tìm GTNN
\(C=\dfrac{\left(x-y\right)^2}{xy}+\dfrac{6xy}{\left(x+y\right)^2}\)
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NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 3 2021
\(A=\dfrac{x^2+y^2}{xy}+\dfrac{xy}{x^2+y^2}=\dfrac{x^2+y^2}{4xy}+\dfrac{xy}{x^2+y^2}+\dfrac{3\left(x^2+y^2\right)}{4xy}\)
\(A\ge2\sqrt{\dfrac{\left(x^2+y^2\right)xy}{4xy\left(x^2+y^2\right)}}+\dfrac{3.2xy}{4xy}=\dfrac{5}{2}\)
Dấu "=" xảy ra khi \(x=y\)
\(C=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{6xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{6xy}{\left(x+y\right)^2}-4\)
\(C=\dfrac{3\left(x+y\right)^2}{8xy}+\dfrac{6xy}{\left(x+y\right)^2}+\dfrac{5\left(x+y\right)^2}{8xy}-4\)
\(C\ge2\sqrt{\dfrac{18xy\left(x+y\right)^2}{8xy\left(x+y\right)^2}}+\dfrac{5.4xy}{8xy}-4=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y\)
14 tháng 1 2021
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...
HN
17 tháng 1 2021
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.
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 12 2020
\(\dfrac{\left(x+y+1\right)^2}{xy+x+y}\ge\dfrac{3\left(xy+x+y\right)}{xy+x+y}=3\)
\(\Rightarrow A=\dfrac{8\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{xy+x+y}{\left(x+y+1\right)^2}\)
\(A\ge\dfrac{8}{9}.3+2\sqrt{\dfrac{\left(x+y+1\right)^2\left(xy+x+y\right)}{\left(xy+x+y\right)\left(x+y+1\right)^2}}=\dfrac{10}{3}\)
Dấu "=" xảy ra khi \(x=y=1\)
=t (t>=3). cho dễ làm
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 4 2022
\(A=\dfrac{x^2+y^2}{xy}+\dfrac{2xy}{x^2+y^2}=\dfrac{x^2+y^2}{2xy}+\dfrac{x^2+y^2}{2xy}+\dfrac{2xy}{x^2+y^2}\)
\(A\ge\dfrac{2xy}{2xy}+2\sqrt{\left(\dfrac{x^2+y^2}{2xy}\right)\left(\dfrac{2xy}{x^2+y^2}\right)}=3\)
Dấu "=" xảy ra khi \(x=y\)
\(B=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{4xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{4xy}{\left(x+y\right)^2}-4\)
\(B=\dfrac{\left(x+y\right)^2}{4xy}+\dfrac{4xy}{\left(x+y\right)^2}+\dfrac{3}{4}.\dfrac{\left(x+y\right)^2}{xy}-4\)
\(B\ge2\sqrt{\dfrac{\left(x+y\right)^2.4xy}{4xy.\left(x+y\right)^2}}+\dfrac{3}{4}.\dfrac{4xy}{xy}-4=1\)
\(B_{min}=1\) khi \(x=y\)
VD
1
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2021
\(\dfrac{x^3}{4\left(y+2\right)}+\dfrac{x\left(y+2\right)}{16}\ge\dfrac{x^2}{4}\) ; \(\dfrac{y^3}{4\left(x+2\right)}+\dfrac{y\left(x+2\right)}{16}\ge\dfrac{y^2}{4}\)
\(\Rightarrow Q+\dfrac{2xy+2x+2y}{16}\ge\dfrac{x^2+y^2}{4}\ge\dfrac{\left(x+y\right)^2}{8}\)
\(\Rightarrow Q\ge\dfrac{\left(x+y\right)^2-\left(x+y\right)}{8}-\dfrac{1}{2}=\dfrac{\left(x+y-4\right)^2+7\left(x+y\right)-16}{8}-\dfrac{1}{2}\)
\(\Rightarrow Q\ge\dfrac{7\left(x+y\right)-16}{8}-\dfrac{1}{2}\ge\dfrac{14\sqrt{xy}-16}{8}-\dfrac{1}{2}=1\)
\(Q_{min}=1\) khi \(x=y=2\)
BC
1
\(C=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{6xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{6xy}{\left(x+y\right)^2}-4\)
\(C=\dfrac{3\left(x+y\right)^2}{8xy}+\dfrac{6xy}{\left(x+y\right)^2}+\dfrac{5\left(x+y\right)^2}{8xy}-4\)
\(C\ge2\sqrt{\dfrac{18xy\left(x+y\right)^2}{8xy\left(x+y\right)^2}}+\dfrac{5.4xy}{8xy}-4=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y\)