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Lời giải:
Từ \(x+y+z=xyz\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)
Đặt \((\frac{1}{a}, \frac{1}{b}, \frac{1}{c})=(x,y,z)\), trong đó $a,b,c>0$ thì ta có:
\(ab+bc+ac=1\) và cần phải CMR:
\(P=\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}+\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}+\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}\)
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Ta có:
\(\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}=\sqrt{(b^2+1)(c^2+1)}-b\sqrt{c^2+1}-c\sqrt{b^2+1}\)
\(=\sqrt{(b^2+ab+bc+ac)(c^2+ac+bc+ab)}-b\sqrt{c^2+ac+bc+ab}-c\sqrt{b^2+ab+bc+ac}\)
\(=\sqrt{(b+a)(b+c)(c+a)(c+b)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}\)
\(=(b+c)\sqrt{(a+b)(a+c)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}(1)\)
Tương tự:
\(\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}=(a+c)\sqrt{(b+a)(b+c)}-a\sqrt{(c+a)(c+b)}-c\sqrt{(a+b)(a+c)}(2)\)
\(\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}=(a+b)\sqrt{(c+a)(c+b)}-b\sqrt{(a+b)(a+c)}-a\sqrt{(b+c)(b+a)}(3)\)
Từ \((1);(2);(3)\Rightarrow P=(b+c-c-b)\sqrt{(a+b)(a+c)}+(a+c-c-a)\sqrt{(b+a)(b+c)}+(a+b-b-a)\sqrt{(c+a)(c+b)}\)
\(=0\)
Ta có đpcm.
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:
\(\sum \sqrt{\frac{(x^2+1)(y^2+1)}{z^2+1}}=\sum \sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}=\sum \sqrt{(x+y)^2}\)
$=\sum (x+y)=2(x+y+z)$
Lời giải:
Ta có
\(xy+yz+xz=1\Rightarrow x^2+1=x^2+xy+yz+xz=(x+y)(x+z)\)
Tương tự: \(\left\{\begin{matrix} y^2+1=(y+z)(y+x)\\ z^2+1=(z+x)(z+y)\end{matrix}\right.\)
Do đó \(A=x\sqrt{\frac{(y+z)(y+x)(x+z)(z+y)}{(x+y)(x+z)}}+y\sqrt{\frac{(z+x)(z+y)(x+y)(x+z)}{(y+z)(y+x)}}+z\sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}\)
\(\Leftrightarrow A=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2\)
Vậy \(A=2\)
\(\left(\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}\right)\rightarrow\left(a,b,c\right)\Rightarrow\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=1\end{matrix}\right.\)
Và \(Q=\sqrt{\dfrac{bc}{a^2+1}}+\sqrt{\dfrac{ab}{c^2+1}}+\sqrt{\dfrac{ca}{b^2+1}}\)
\(=\sqrt{\dfrac{bc}{a^2+ab+bc+ca}}+\sqrt{\dfrac{ab}{c^2+ab+bc+ca}}+\sqrt{\dfrac{ca}{b^2+ab+bc+ca}}\)
\(=\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{a+c}{a+c}+\dfrac{b+c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" <=> \(a=b=c=\dfrac{1}{\sqrt{3}}\Leftrightarrow x=y=z=\sqrt{3}\)
Lời giải:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\Leftrightarrow x+y+z=xyz\)
\(\Rightarrow x(x+y+z)=x^2yz\)
\(\Rightarrow x(x+y+z)+yz=x^2yz+yz\Leftrightarrow (x+y)(x+z)=yz(1+x^2)\)
Do đó: \(\frac{x}{\sqrt{yz(x^2+1)}}=\frac{x}{\sqrt{(x+y)(x+z)}}\). Tương tự với các phân thức còn lại suy ra:
\(Q=\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(y+z)(y+x)}}+\frac{z}{\sqrt{(z+x)(z+y)}}\)
Áp dụng BĐT AM-GM ta có:
\(Q\leq \frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)+\frac{1}{2}\left(\frac{y}{y+z}+\frac{y}{y+x}\right)+\frac{1}{2}\left(\frac{z}{z+x}+\frac{z}{z+y}\right)\)
\(\Leftrightarrow Q\leq \frac{1}{2}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{3}{2}\)
Vậy \(Q_{\max}=\frac{3}{2}\Leftrightarrow x=y=z=\sqrt{3}\)
1 + y2 = xy + yz + xz + y2 = (x + y)(y + z)
1 + z2 = xy + yz + xz + z2 = (x + z)(z + y)
1 + x2 = xy + yz + xz + x2 = (y + x)(x + z)
Sau khi thay vào và rút gọn ta được
S = x(y + z) + y(x + z) + z(x + y)
S = 2(xy + yz + xz) = 2.1 = 2
3) Gợi ý: Thay 1=xy+yz+xz
\(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=x\sqrt{\dfrac{\left(y^2+xy+yz+xz\right)\left(z^2+xy+yz+xz\right)}{x^2+xy+yz+xz}}=x\sqrt{\dfrac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+z\right)\left(x+y\right)}}=x\sqrt{\left(y+z\right)^2}=x\left(y+z\right)\)
Tương tự rồi cộng vào
Lời giải:
Vì $xy+yz+xz=1$ nên:
\(x^2+1=x^2+xy+yz+xz=x(x+y)+z(x+y)=(x+z)(x+y)\)
\(y^2+1=y^2+xy+yz+xz=y(y+x)+z(y+x)=(y+z)(y+x)\)
\(z^2+1=z^2+xy+yz+xz=(z^2+xz)+(xy+yz)=z(z+x)+y(x+z)=(z+y)(z+x)\)
Do đó:
\(P=x\sqrt{\frac{(y+z)(y+x)(z+x)(z+y)}{(x+y)(x+z)}}+y\sqrt{\frac{(z+x)(z+y)(x+y)(x+z)}{(y+x)(y+z)}}+z\sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}\)
\(=x\sqrt{(y+z)^2}+y\sqrt{(x+z)^2}+z\sqrt{(x+y)^2}\)
\(=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2\)
Ta có 1+x2 = xy + yz + xz +x2 = ( x+ z)(x+y)
TT : 1+y2 = (y+z)(y+x)
1+z2 = (z+x)(z+y)
⇒ P = 2
Vậy P =2
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 \(x=y=z=\sqrt{\frac{1}{672}}\)
cộng 2016 nhé