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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 \(x=y=z=\sqrt{\frac{1}{672}}\)
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 \(A=2\)
Lời giải:
Ta có: \(xy+yz+xz=1\)
\(\Rightarrow \left\{\begin{matrix} x^2+1=x^2+xy+yz+xz=(x+y)(x+z)\\ y^2+1=y^2+xy+yz+xz=(y+z)(y+x)\\ z^2+1=z^2+xy+yz+xz=(z+x)(z+y)\end{matrix}\right.\)
Do đó:
\(\sqrt{\frac{(y^2+1)(z^2+1)}{x^2+1}}=\sqrt{\frac{(y+z)(y+x)(z+x)(z+y)}{(x+y)(x+z)}}=\sqrt{(y+z)^2}=y+z\)
\(\Rightarrow x\sqrt{\frac{(y^2+1)(z^2+1)}{x^2+1}}=x(y+z)\)
Hoàn toàn tt:
\(y\sqrt{\frac{(z^2+1)(x^2+1)}{y^2+1}}=y(x+z)\); \(z\sqrt{\frac{(x^2+1)(y^2+1)}{z^2+1}}=z(x+y)\)
Do đó:
\(A=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2\)
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.
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
a) Ta có : \(1+x^2=xy+yz+zx+x^2=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(z+x\right)\)
b) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)
\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)
Ta có:
\(x^2+1=x^2+xy+yz+zx\)
\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
Tương tự:
\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)
\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
TH1: x,y,z <0
\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)
TH2: x,y,z>0
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)
Ta có \(1+z^2=xy+yz+zx+z^2\)
\(=y\left(x+z\right)+z\left(x+z\right)\)
\(=\left(x+z\right)\left(y+z\right)\)
CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)
Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)
Tương tự như thế, ta được
\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.
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:
\(P=x\sqrt{\frac{(y+z)(y+x)(z+x)(z+y)}{(x+y)(x+z)}}+y\sqrt{\frac{(z+x)(z+y)(x+y)(x+z)}{(y+x)(y+z)}}+z\sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}\)
\(=x\sqrt{(y+z)^2}+y\sqrt{(x+z)^2}+z\sqrt{(x+y)^2}\)
\(=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2\)
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:
\(\sum \sqrt{\frac{(x^2+1)(y^2+1)}{z^2+1}}=\sum \sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}=\sum \sqrt{(x+y)^2}\)
$=\sum (x+y)=2(x+y+z)$