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25 tháng 7 2019

\(P=\frac{x\left(\sqrt{y}-\sqrt{z}\right)-y\left(\sqrt{x}-\sqrt{z}\right)+z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}\)

\(P=\frac{x\left(\sqrt{y}-\sqrt{z}\right)-y\left[\left(\sqrt{y}-\sqrt{z}\right)+\left(\sqrt{x}-\sqrt{y}\right)\right]+z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}\)

\(P=\frac{\left(x-y\right)\left(\sqrt{y}-\sqrt{z}\right)+\left(z-y\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}\)

\(P=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)-\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}\)

\(P=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left[\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{y}+\sqrt{z}\right)\right]}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}\)

\(P=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}=1\)

=> đpcm

\(P=\dfrac{x\left(\sqrt{y}-\sqrt{z}\right)-y\left(\sqrt{x}-\sqrt{z}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}+\dfrac{z}{\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{x\sqrt{y}-x\sqrt{z}-y\sqrt{x}+y\sqrt{z}+z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{z}\left(x-y\right)+z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{xy}-\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)+z\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{\left(\sqrt{xy}-\sqrt{zx}-\sqrt{zy}+z\right)}{\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{y}-\sqrt{z}\right)-\sqrt{z}\left(\sqrt{y}-\sqrt{z}\right)}{\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}{\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

=1

4 tháng 8 2017

phân số thứ 3 sai

14 tháng 7 2019

mình cần gấp, thanks các bạn

14 tháng 7 2019

Đề chắc chắn đúng chứ bạn??

29 tháng 8 2019

\(\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)}+\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{x}\right)}+\)\(\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{z}-\sqrt{y}\right)}\)

\(=-\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{z}-\sqrt{x}\right)}-\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(-\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\frac{-x\left(\sqrt{y}-\sqrt{z}\right)-y\left(\sqrt{z}-\sqrt{x}\right)-z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

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

\(=\frac{-\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{z}\left(x-y\right)-z\left(\sqrt{x}-y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{-\sqrt{xy}+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)-z}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

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

\(=\frac{\sqrt{y}\left(\sqrt{z}-\sqrt{x}\right)-\sqrt{z}\left(\sqrt{z}-\sqrt{x}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(=\frac{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)

\(A=\frac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}+\frac{\left(x+z\right)\sqrt{\left(x+y\right)\left(y+z\right)}}{y}+\frac{\left(x+y\right)\sqrt{\left(y+z\right)\left(x+z\right)}}{z}.\)

Áp dụng bất đẳng thức Bunhiacopski ta có

\(\left(x+y\right)\left(x+z\right)\ge\left(x+\sqrt{yz}\right)^2\)

Tương tự \(\left(x+y\right)\left(y+z\right)\ge\left(y+\sqrt{xz}\right)^2\)

                 \(\left(y+z\right)\left(x+z\right)\ge\left(z+\sqrt{xy}\right)^2\)

\(\Rightarrow A\ge\frac{\left(y+z\right)\left(x+\sqrt{yz}\right)}{x}+\frac{\left(x+z\right)\left(y+\sqrt{xz}\right)}{y}+\frac{\left(x+y\right)\left(z+\sqrt{xy}\right)}{z}\)

hay \(A\ge2\left(x+y+z\right)+\frac{\sqrt{yz}\left(y+z\right)}{x}+\frac{\left(x+z\right)\sqrt{xz}}{y}+\frac{\left(x+y\right)\sqrt{xy}}{z}\)

\(\Leftrightarrow A\ge2\left(x+y+z\right)+\frac{yz\sqrt{yz}\left(y+z\right)}{xyz}+\frac{xz\sqrt{xz}\left(x+z\right)}{xyz}+\frac{xy\sqrt{xy}\left(x+y\right)}{xyz}\)

Đặt \(M=\frac{yz\sqrt{yz}\left(y+z\right)}{xyz}+\frac{xz\sqrt{xz}\left(x+z\right)}{xyz}+\frac{xy\sqrt{xy}\left(x+y\right)}{xyz}\)

Ta có \(\left(x,y,z\right)\rightarrow\left(a^2,b^2,c^2\right)\)

Khi đó \(M=\frac{a^3b^3\left(a^2+b^2\right)+b^3c^3\left(b^2+c^2\right)+c^3a^3\left(a^2+c^2\right)}{a^2b^2c^2}\)

ÁP DỤNG BĐT AM-GM ta có

\(a^5b^3+a^3b^5\ge2\sqrt{a^8b^8}=2a^4b^4\)

\(b^5c^3+b^3c^5\ge2\sqrt{b^8c^8}=2b^4c^4\)

\(a^5c^3+a^3c^5\ge2\sqrt{a^8c^8}=2a^4c^4\)

Cộng từng vế ta được 

\(a^3b^3\left(a^2+b^2\right)+b^3c^3\left(b^2+c^2\right)+c^3a^3\left(a^2+c^2\right)\ge2\left(a^4b^4+b^4c^4+c^4a^4\right)\)

              \(\ge2a^2b^2c^2\left(a^2+b^2+c^2\right)\)

\(\Rightarrow M\ge2\left(a^2+b^2+c^2\right)=2\left(x+y+z\right)\)

\(\Rightarrow A\ge4\left(x+y+z\right)=4\sqrt{2019}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{\sqrt{2019}}{3}\)

18 tháng 10 2020

Vì xyz=1\(\Rightarrow x^2\left(y+z\right)\ge2x^2\sqrt{yz}=2x\sqrt{x}\)

Tương tự \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2=\left(x+y\right)\ge2z\sqrt{z}\)

\(\Rightarrow P\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)

Đặt \(x\sqrt{x}+2y\sqrt{y}=a;y\sqrt{y}+2z\sqrt{z}=b;z\sqrt{z}+2x\sqrt{x}=c\)

\(\Rightarrow x\sqrt{x}=\frac{4c+a-2b}{9};y\sqrt{y}=\frac{4a+b-2c}{9};z\sqrt{z}=\frac{4b+c-2a}{9}\)

\(\Rightarrow P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{a}+\frac{4b+c-2a}{b}\right)\)

\(=\frac{2}{9}\text{ }\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\ge\frac{2}{9}\left(4.3+2-6\right)=2\)

Min P =2 khi và chỉ khi a=b=c khi va chỉ khi x=y=z=1

12 tháng 10 2016

Đặt \(\sqrt{x}=a\) , \(\sqrt{y}=b\) , \(\sqrt{z}=c\)

Suy ra \(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)

\(=-\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Xét tử : \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left[-\left(a-b\right)-\left(c-a\right)\right]+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(c^2-a^2\right)+\left(c-a\right)\left(b^2-a^2\right)=\left(a-b\right)\left(c-a\right)\left(c+a\right)+\left(c-a\right)\left(b-a\right)\left(b+a\right)\)

\(=\left(a-b\right)\left(c-a\right)\left(c+a-a-b\right)=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)

Suy ra \(P=-\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)