Cho ba số thực dương x,y,z. Biểu thức P=\(\dfrac{1}{2}\left(x^2+y^2+z^2\right)+\dfrac{x}{yz}+\dfrac{y}{zx}+\dfrac{z}{xy}\) có GTNN là bao nhiêu
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NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1 2021
Ta có:
\(\dfrac{x}{yz}+\dfrac{y}{zx}+\dfrac{z}{xy}=\dfrac{1}{2}\left(\dfrac{x}{yz}+\dfrac{y}{zx}+\dfrac{x}{yz}+\dfrac{z}{xy}+\dfrac{y}{zx}+\dfrac{z}{xy}\right)\ge\dfrac{1}{2}\left(\dfrac{2}{z}+\dfrac{2}{y}+\dfrac{2}{x}\right)\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(x^2+y^2+z^2\right)+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(x^2+\dfrac{1}{x}+\dfrac{1}{x}\right)+\dfrac{1}{2}\left(y^2+\dfrac{1}{y}+\dfrac{1}{y}\right)+\dfrac{1}{2}\left(z^2+\dfrac{1}{z}+\dfrac{1}{z}\right)\)
\(\Rightarrow P\ge\dfrac{3}{2}\sqrt[3]{\dfrac{x^2}{x^2}}+\dfrac{3}{2}\sqrt[3]{\dfrac{y^2}{y^2}}+\dfrac{3}{2}\sqrt[3]{\dfrac{z^2}{z^2}}=\dfrac{9}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 4 2021
\(VT\ge3\sqrt[3]{\dfrac{x^3y^3z^3\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}}=3xyz\) (dpcm)
AH
Akai Haruma
Giáo viên
13 tháng 11 2023
Lời giải:
$xy+yz+xz=1$
$\Rightarrow x^2+1=x^2+xy+yz+xz=(x+y)(x+z)$
Tương tự: $y^2+1=(y+z)(y+x); z^2+1=(z+x)(z+y)$
Khi đó:
\(\sum \sqrt{\frac{(x^2+1)(y^2+1)}{z^2+1}}=\sum \sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}=\sum \sqrt{(x+y)^2}\)
$=\sum (x+y)=2(x+y+z)$
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 12 2021
\(P=\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{z}\right)\ge\dfrac{1}{y}.\dfrac{4}{x+z}=\dfrac{4}{y\left(x+z\right)}\ge\dfrac{4}{\dfrac{\left(y+x+z\right)^2}{4}}=4\)
\(P_{min}=4\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{2};1;\dfrac{1}{2}\right)\)
ND
Nguyễn Đức Trí
VIP
20 tháng 8 2023
Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)
\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)
\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)
Tương tự ta được
\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)
\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :
\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)
\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)
\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 12 2020
\(A\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{1}{2}\left(x+y+z\right)\ge\dfrac{1}{2}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=\dfrac{1}{2}\)
\(A_{min}=\dfrac{1}{2}\) khi \(x=y=z=\dfrac{1}{3}\)
\(P=\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{z^2}{2}+\dfrac{x^2+y^2+z^2}{xyz}\)
\(\Rightarrow P\ge\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{z^2}{2}+\dfrac{xy+xz+yz}{xyz}\)
\(\Rightarrow P\ge\dfrac{x^2}{2}+\dfrac{1}{x}+\dfrac{y^2}{2}+\dfrac{1}{y}+\dfrac{z^2}{2}+\dfrac{1}{z}\)
\(\Rightarrow P\ge\left(\dfrac{x^2}{2}+\dfrac{1}{2x}+\dfrac{1}{2x}\right)+\left(\dfrac{y^2}{2}+\dfrac{1}{2y}+\dfrac{1}{2y}\right)+\left(\dfrac{z^2}{2}+\dfrac{1}{2z}+\dfrac{1}{2z}\right)\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{x^2}{2}.\dfrac{1}{2x}.\dfrac{1}{2x}}+3\sqrt[3]{\dfrac{y^2}{2}.\dfrac{1}{2y}.\dfrac{1}{2y}}+3\sqrt[3]{\dfrac{z^2}{2}.\dfrac{1}{2z}.\dfrac{1}{2z}}=\dfrac{9}{2}\)
\(\Rightarrow P_{min}=\dfrac{9}{2}\) khi \(x=y=z=1\)