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22 tháng 8 2020

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

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

21 tháng 8 2020

Bài này phải tìm GTLN chứ nhỉ?!

12 tháng 11 2018

\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)

\(=\frac{1}{\sqrt{\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y-z\right)^2+\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z-x\right)^2+\frac{1}{2}\left(z^2+x^2\right)}}\)

\(\le\frac{1}{\sqrt{\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z^2+x^2\right)}}\)

\(\le\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

14 tháng 8 2020

\(P=\frac{\sqrt{1+x^2+y^2}}{xy}+\frac{\sqrt{1+y^2+z^2}}{yz}+\frac{\sqrt{1+z^2+x^2}}{zx}\)

\(\ge\text{Σ}\frac{\sqrt{\frac{\left(1+x+y\right)^2}{3}}}{xy}\text{=}\frac{1+x+y}{xy\sqrt{3}}\)

\(=\frac{\sqrt{3}}{3}\left(\frac{1+x+y}{xy}+\frac{1+y+z}{yz}+\frac{1+z+x}{zx}\right)\)

\(=\frac{\sqrt{3}}{3}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{x}\right)\)

\(=\frac{\sqrt{3}}{3}\left(x+y+z+2xy+2yz+2zx\right)\)\(\ge\frac{\sqrt{3}}{3}\left(3\sqrt[3]{xyz}+2\cdot3\sqrt[3]{x^2y^2z^2}\right)=\frac{\sqrt{3}}{3}\left(3+6\right)=3\sqrt{3}\)

Dấu = xảy ra khi \(x=y=z=1\)

14 tháng 8 2020

\(P=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}+\frac{3}{x+y+z}\)

\(=x+y+z+\frac{9}{x+y+z}-\frac{6}{x+y+z}\)

\(\ge6-\frac{6}{3\sqrt[3]{xyz}}=6-\frac{6}{3}=4\)

Dấu = xảy ra khi x = y = z = 1