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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 \(x=y=z=\sqrt{\frac{1}{672}}\)
\(\sqrt{2x\left(y+z\right)}< =\dfrac{2x+y+z}{2}\)
=>\(\dfrac{1}{\sqrt{x\left(y+z\right)}}>=\dfrac{2\sqrt{2}}{2x+y+z}\)
=>\(P>=2\sqrt{2}\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)
\(\Leftrightarrow P>=2\sqrt{2}\cdot\dfrac{\left(1+1+1\right)^2}{\left(2x+y+z\right)+x+2y+z+x+y+2z}=\dfrac{18\sqrt{2}}{4\cdot18\sqrt{2}}=\dfrac{1}{4}\)
Dấu = xảy ra khi x=y=z=6căn 2
Áp dụng BĐT AM-GM ta có:
\(\frac{\left(x+1\right)\left(y+1\right)^2}{3\sqrt[3]{x^2y^2}+1}\ge\frac{\left(x+1\right)\left(y+1\right)^2}{xy+x+y+1}=\frac{\left(x+1\right)\left(y+1\right)^2}{\left(x+1\right)\left(y+1\right)}=y+1\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(P\ge x+y+z+3=6\)
Dấu "=" <=> x=y=z=1
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.
Áp dụng bđt Cauchy-Schwarz:
\(A=\dfrac{1}{\sqrt{x\left(y+2z\right)}}+\dfrac{1}{\sqrt{y\left(z+2x\right)}}+\dfrac{1}{\sqrt{z\left(x+2y\right)}}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}}\)
\(=\dfrac{9}{\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}}\)
Áp dụng liên tiếp Bunyakovsky và AM-GM:
\(\left(\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left[x\left(y+2z\right)+y\left(z+2x\right)+z\left(x+2y\right)\right]\)
\(=3.3\left(xy+yz+xz\right)\)
Mà \(3\left(xy+yz+xz\right)\le\left(x+y+z\right)^2=3\)
\(3.3\left(xy+yz+xz\right)\le3.3=9\)
\(\Leftrightarrow\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+z\sqrt{\left(x+2y\right)}\le\sqrt{9}=3\)
\(\Leftrightarrow A\ge\dfrac{9}{3}=3."="\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)
Áp dụng bất đẳng thức cauchy:
\(P=\sum\dfrac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}\ge\sum\dfrac{2x^2\sqrt{yz}}{y\sqrt{y}+2z\sqrt{z}}=\sum\dfrac{2\sqrt{x^3}\sqrt{xyz}}{\sqrt{y^3}+2\sqrt{z^3}}=\sum\dfrac{2\sqrt{x^3}}{\sqrt{y^3}+2\sqrt{z^3}}\)(vì xyz=1).
đặt \(\left\{{}\begin{matrix}\sqrt{x^3}=a\\\sqrt{y^3}=b\\\sqrt{z^3}=c\end{matrix}\right.\)(\(a,b,c>0\))thì giả thiết trở thành cho abc=1. tìm Min \(P=\dfrac{2a}{b+2c}+\dfrac{2b}{c+2a}+\dfrac{2c}{a+2b}\)
Áp dụng BĐT cauchy-schwarz:
\(P=2\left(\dfrac{a^2}{ab+2ac}+\dfrac{b^2}{bc+2ab}+\dfrac{c^2}{ac+2bc}\right)\ge\dfrac{2\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=2\)( AM-GM \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\))
Dấu = xảy ra khi a=b=c=1 hay x=y=z=1
Sửa đề \(\dfrac{\left(x+1\right)\left(y+1\right)^2}{3\sqrt[3]{x^2z^2}+1}+\dfrac{\left(y+1\right)\left(z+1\right)^2}{3\sqrt[3]{x^2y}+1}+\dfrac{\left(z+1\right)\left(x+1\right)^2}{3\sqrt[3]{y^2z^2}+1}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{\left(x+1\right)\left(y+1\right)^2}{3\sqrt[3]{x^2z^2}+1}=\dfrac{\left(x+1\right)\left(y+1\right)^2}{3\sqrt[3]{x\cdot z\cdot xz}+1}\ge\dfrac{\left(x+1\right)\left(y+1\right)^2}{x+z+xz+1}\)
\(=\dfrac{\left(x+1\right)\left(y+1\right)^2}{\left(x+1\right)\left(z+1\right)}=\dfrac{\left(y+1\right)^2}{z+1}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{\left(y+1\right)\left(z+1\right)^2}{3\sqrt[3]{x^2y^2}+1}\ge\dfrac{\left(z+1\right)^2}{x+1};\dfrac{\left(z+1\right)\left(x+1\right)^2}{3\sqrt[3]{y^2z^2}+1}\ge\dfrac{\left(x+1\right)^2}{y+1}\)
Cộng theo vế 3 BĐT trên rồi áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT\ge\dfrac{\left(x+y+z+3\right)^2}{x+y+z+3}=x+y+z+3=VP\)
sửa đề rồi mà vẫn sai :v