\(\ne0\) thỏa mãn x + y + z = 0 .

CMR : \(\sqr...">

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25 tháng 7 2018

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

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

Vì x+y+z =0 \(\Rightarrow A=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\) (đpcm)

AH
Akai Haruma
Giáo viên
14 tháng 8 2018

Bài 1:

\((x,y,z)=(\frac{2a^2}{bc}; \frac{2b^2}{ca}; \frac{2c^2}{ab})\) (\(a,b,c>0\) )

Khi đó:

\(\text{VT}=\frac{\frac{4a^4}{b^2c^2}}{\frac{4a^4}{b^2c^2}+\frac{4a^2}{bc}+1}+\frac{\frac{4b^4}{c^2a^2}}{\frac{4b^4}{c^2a^2}+\frac{4b^2}{ca}+4}+\frac{\frac{4c^4}{a^2b^2}}{\frac{4c^4}{a^2b^2}+\frac{4c^2}{ab}+4}\)

\(=\frac{a^4}{a^4+a^2bc+b^2c^2}+\frac{b^4}{b^4+b^2ac+a^2c^2}+\frac{c^4}{c^4+c^2ab+a^2b^2}\)

\(\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+a^2bc+b^2ac+c^2ab+(a^2b^2+b^2c^2+c^2a^2)}\)

(Áp dụng BĐT Cauchy_Schwarz)

Theo BĐT Cauchy dễ thấy:

\(a^2b^2+b^2c^2+c^2a^2\geq a^2bc+b^2ca+c^2ab\)

\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)}=\frac{(a^2+b^2+c^2)^2}{(a^2+b^2+c^2)^2}=1\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=2$

AH
Akai Haruma
Giáo viên
14 tháng 8 2018

Bài 2:

Đặt \((x,y,z)=\left(\frac{a}{b};\frac{b}{c}; \frac{c}{a}\right)\)

Ta có:

\(\text{VT}=\left(\frac{a}{b}+\frac{c}{b}-1\right)\left(\frac{b}{c}+\frac{a}{c}-1\right)\left(\frac{c}{a}+\frac{b}{a}-1\right)\)

\(=\frac{(a+c-b)(b+a-c)(c+b-a)}{abc}\)

Áp dụng BĐT Cauchy:

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

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

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

Nhân theo vế:

\(\Rightarrow [(a+c-b)(b+a-c)(c+b-a)]^2\leq (abc)^2\)

\(\Rightarrow (a+c-b)(b+a-c)(c+b-a)\leq abc\)

\(\Rightarrow \text{VT}\leq 1\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=1$

26 tháng 7 2018

\(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{xz}-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{xz}}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2.\dfrac{x+y+z}{xyz}}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\text{|}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\text{|}\)

26 tháng 7 2018

\(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)-2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{z}{xyz}+\dfrac{y}{xyz}+\dfrac{x}{xyz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)

24 tháng 5 2018

Ta có BĐT:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

\(\Leftrightarrow6\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)+2016\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow7.\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le2016\)
Xét \(P=\frac{1}{\sqrt{3\left(2x^2+y^2\right)}}+\frac{1}{\sqrt{3\left(2y^2+z^2\right)}}+\frac{1}{\sqrt{3\left(2z^2+x^2\right)}}\)
\(P^2=\left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2x^2+y^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2y^2+z^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2z^2+x^2}}\right)^2\)
Áp dụng BĐT Bunhiacopxki ta có:
\(P^2\le\left(\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2\right)\left(\left(\frac{1}{\sqrt{2x^2+y^2}}\right)^2+\left(\frac{1}{\sqrt{2y^2+z^2}}\right)^2+\left(\frac{1}{\sqrt{2z^2+x^2}}\right)^2\right)\)
\(\Leftrightarrow P^2\le\frac{1}{2x^2+y^2}+\frac{1}{2y^2+z^2}+\frac{1}{2z^2+x^2}\)
Mặt khác ta có:
\(\frac{1}{2x^2+y^2}=\frac{1}{x^2+x^2+y^2}\le\frac{1}{9}\left(\frac{1}{x^2}+\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\frac{1}{2y^2+z^2}\le\frac{1}{9}\left(\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
\(\frac{1}{2z^2+x^2}\le\frac{1}{9}\left(\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}\right)\)
\(\Rightarrow P^2\le\frac{1}{3}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le\frac{1}{3}.2016=672\)
\(\Rightarrow P\le4\sqrt{42}\)
Dấu '=' xảy ra khi \(x=y=z=\sqrt{\frac{1}{672}}\)
 

23 tháng 5 2018

cộng 2016 nhé

14 tháng 7 2018

Bài 1 :

Ta có : \(\dfrac{1}{3a^2+b^2}+\dfrac{2}{b^2+3ab}=\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\)

Theo BĐT Cô - Si dưới dạng engel ta có :

\(\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\ge\dfrac{\left(1+2\right)^2}{3a^2+6ab+3b^2}=\dfrac{9}{3\left(a+b\right)^2}=\dfrac{9}{3.1}=3\)

Dấu \("="\) xảy ra khi : \(a=b=\dfrac{1}{2}\)

2 tháng 12 2019

Ta có:\(\frac{4+4\sqrt{1+x^2}}{4x}\le\frac{4+5+x^2}{4x}=\)\(\frac{x^2+9}{4x}\)Tương tự ta đc P\(\le\frac{x+y+z}{4}+\frac{9}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=\frac{1}{4}\left(x+y+z\right)+\frac{9}{4}\left(\frac{xy+yz+zx}{xyz}\right)\)\(\le\frac{1}{4}\left(x+y+z\right)+\frac{9}{4}\cdot\frac{\left(x+y+z\right)^2}{3\left(x+y+z\right)}\)\(=x+y+z\)

Dấu '='xảy ra <=>\(\hept{\begin{cases}x+y+z=xyz\\x=y=z\end{cases}\Rightarrow x=y=z=}\)\(\frac{1}{\sqrt{3}}\)