Cho x y z > 0 và xyz=1. Tìm GTNN của \(P=\frac{1}{x+y+z}-\frac{2}{xy+yz+zx}\)
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TN
19 tháng 7 2017
Áp dụng BĐT AM-GM ta có:
\(\frac{\sqrt{1+x^3+y^3}}{xy}\ge\frac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}=\frac{\sqrt{3xy}}{xy}=\frac{\sqrt{3}}{\sqrt{xy}}\)
Tương tự cho 2 BĐT còn lại ta có:
\(\frac{\sqrt{1+y^3+z^3}}{yz}\ge\frac{\sqrt{3}}{\sqrt{yz}};\frac{\sqrt{1+z^3+x^3}}{xz}\ge\frac{\sqrt{3}}{\sqrt{xz}}\)
Cộng theo vế 3 BĐT trên ta có:
\(M\ge\sqrt{3}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)=\sqrt{3}\cdot\left(\frac{\sqrt{x}}{\sqrt{xyz}}+\frac{\sqrt{y}}{\sqrt{xyz}}+\frac{\sqrt{z}}{\sqrt{xyz}}\right)\)
\(=\sqrt{3}\cdot\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{xyz}}\ge\sqrt{3}\cdot\frac{3\sqrt[3]{\sqrt{xyz}}}{1}=3\sqrt{3}\)
Khi \(x=y=z=1\)
30 tháng 6 2020
Ta có: \(xy+yz+zx=xyz\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)ta có: \(a,b,c>0;a+b+c=1\)do đó 0<a,b,c<1
\(P=\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+6\left(ab+bc+ca\right)\)
\(=\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+2\left(a+b+c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)
\(=\left(\frac{b^2}{a}-2b+a\right)+\left(\frac{c^2}{b}-2c+b\right)+\left(\frac{a^2}{c}-2a+c\right)-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)
\(=\frac{\left(a-b\right)^2}{a}+\frac{\left(b-c\right)^2}{b}+\frac{\left(c-a\right)^2}{c}-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)
\(=\frac{\left(1-a\right)\left(a-b\right)^2}{a}+\frac{\left(1-b\right)\left(b-c\right)^2}{b}+\frac{\left(1-c\right)\left(c-a\right)^2}{c}+3\ge3\)
Vậy GTNN của P=3
12 tháng 11 2018
\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)
\(=\frac{1}{\sqrt{\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y-z\right)^2+\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z-x\right)^2+\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{1}{\sqrt{\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
ND
0
Dat \(\left(a,b,c\right)=\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\left(a,b,c>0,abc=1\right)\)
Ta co \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow\frac{3}{ab+bc+ca}\ge\frac{9}{\left(a+b+c\right)^2}\left(1\right)\)
BDT phu \(1+\frac{3}{ab+bc+ca}\ge\frac{6}{a+b+c}\left(2\right)\)
Do (1) nen (2) tuong duong voi
\(1+\frac{9}{\left(a+b+c\right)^2}\ge\frac{6}{a+b+c}\Leftrightarrow\left(1-\frac{3}{a+b+c}\right)^2\ge0\left(dung\right)\)
Suy ra (2) duoc chung minh
Do \(abc=1\Rightarrow\hept{\begin{cases}ab=\frac{1}{xy}=\frac{xyz}{xy}=z\\bc=x\\ca=y\end{cases}}\)
nen (2) tuong duong \(1+\frac{3}{x+y+z}\ge\frac{6}{xy+yz+zx}\)
=> \(\frac{1}{x+y+z}\ge\frac{1}{3}\left(\frac{6}{x+y+z}-1\right)=\frac{2}{x+y+z}-\frac{1}{3}\)
Suy ra \(P\ge\frac{2}{x+y+z}-\frac{1}{3}-\frac{2}{x+y+z}=-\frac{1}{3}\)
Dau = xay ra khi x=y=z=1