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Lời giải:
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\Leftrightarrow xy+yz+xz=xyz\)
\(\Rightarrow x^2+xy+yz+xz=x^2+xyz=x(x+yz)\)
\(\Leftrightarrow x+yz=\frac{x^2+xy+yz+xz}{x}=\frac{(x+y)(x+z)}{x}\)
\(\Rightarrow \sqrt{x+yz}=\sqrt{\frac{(x+y)(x+z)}{x}}\)
Áp dụng BĐT Bunhiacopxky:\((x+y)(x+z)\geq (x+\sqrt{yz})^2\)
\(\Rightarrow \sqrt{x+yz}=\sqrt{\frac{(x+y)(x+z)}{x}}\geq \frac{x+\sqrt{yz}}{\sqrt{x}}\)
Hoàn toàn tương tự:
\(\sqrt{y+xz}\geq \frac{y+\sqrt{xz}}{\sqrt{y}}\); \(\sqrt{z+xy}\geq \frac{z+\sqrt{xy}}{\sqrt{z}}\)
Cộng theo vế các BĐT đã thu được ta có:
\(\text{VT}\geq \frac{x+\sqrt{yz}}{\sqrt{x}}+\frac{y+\sqrt{xz}}{\sqrt{y}}+\frac{z+\sqrt{xy}}{\sqrt{z}}=\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xy+yz+xz}{\sqrt{xyz}}\)
\(\Leftrightarrow \text{VT}\geq \sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xyz}{\sqrt{xyz}}=\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}=\text{VP}\)
Do đó ta có đpcm.
Dấu bằng xảy ra khi \(x=y=z=3\)
Lời giải:
Đặt \((\frac{1}{x}; \frac{1}{y}; \frac{1}{z})=(a,b,c)\). Bài toán trở thành:
Cho $a,b,c>0$ thỏa mãn $a+b+c=1$. CMR:
\(\frac{\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}}{\sqrt{abc}}\geq \sqrt{\frac{1}{abc}}+\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}(*)\)
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Do $a+b+c=1$ nên ta có:
\(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}=\sqrt{a(a+b+c)+bc}+\sqrt{b(a+b+c)}+\sqrt{c(a+b+c)+ab}\)
\(=\sqrt{(a+b)(a+c)}+\sqrt{(b+a)(b+c)}+\sqrt{(c+a)(c+b)}\)
Mà áp dụng BĐT Bunhiacopxky:
\(\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+a)}+\sqrt{(c+a)(c+b)}\geq \sqrt{(a+\sqrt{bc})^2}+\sqrt{(b+\sqrt{ac})^2}+\sqrt{(c+\sqrt{ab})^2}\)
\(=a+\sqrt{bc}+b+\sqrt{ac}+c+\sqrt{ab}=a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
\(1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
Vậy:\(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}\geq 1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
\(\Rightarrow \frac{\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}}{\sqrt{abc}}\geq \sqrt{\frac{1}{abc}}+\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}\)
$(*)$ được cm. BĐT hoàn thành. Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$ hay $x=y=z=3$
Lời giải:
Áp dụng BĐT AM-GM:
\(\sqrt{\frac{xy}{xy+z}}=\sqrt{\frac{xy}{xy+z(x+y+z)}}=\sqrt{\frac{xy}{(z+x)(z+y)}}\leq \frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{z+y}\right)\)
Hoàn toàn tương tự với các phân thức còn lại suy ra:
\(\sum \sqrt{\frac{xy}{xy+z}}\leq \frac{1}{2}\left(\frac{x+z}{x+z}+\frac{y+z}{y+z}+\frac{x+y}{x+y}\right)=\frac{3}{2}\)
Ta có đpcm.
Dấu "=" xảy ra khi $x=y=z=\frac{1}{3}$
Áp dụng BĐT AM-GM, Ta có
\(\sqrt{x-1}\le\dfrac{1+x-1}{2}=\dfrac{x}{2}\Rightarrow yz\sqrt{x-1}\le\dfrac{xyz}{2}\)
Mà \(xz\sqrt{y-2}\le\dfrac{xz\sqrt{2\left(y-2\right)}}{\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\)
\(yx\sqrt{z-3}\le yx.\dfrac{3+z-3}{2\sqrt{3}}=\dfrac{xyz}{2\sqrt{3}}\)
\(\Rightarrow\dfrac{xy\sqrt{x-1}+xz\sqrt{y-2}+yz\sqrt{z-3}}{xyz}\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}=\dfrac{1}{2}+\dfrac{\sqrt{2}}{4}+\dfrac{\sqrt{3}}{6}\)
Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)\rightarrow\left(a;b;c\right)\)\(\Rightarrow\left\{{}\begin{matrix}a+b+c=1\\a;b;c>0\end{matrix}\right.\)
Và \(\dfrac{ab}{\sqrt{a^2+b^2+2c^2}}+\dfrac{bc}{\sqrt{b^2+c^2+2a^2}}+\dfrac{ca}{\sqrt{c^2+a^2+2b^2}}\le\dfrac{1}{2}\)
Ta có:\(\dfrac{ab}{\sqrt{a^2+b^2+2c^2}}=\dfrac{2ab}{\sqrt{\left(1+1+2\right)\left(a^2+b^2+2c^2\right)}}\)
\(\le\dfrac{2ab}{a+b+2c}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{ab+bc}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{bc+ac}{a+b}\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\)
Dấu "=" khi \(a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=\dfrac{1}{9}\)
Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\) \(\ge\) \(\dfrac{2}{\sqrt{xy}}\) (1)
\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\) (2)
\(\dfrac{1}{z}+\dfrac{1}{x}\ge\dfrac{2}{\sqrt{xz}}\) (3)
Cộng (1);(2);(3) vế theo vế ta được:
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)
=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\) (đpcm)
\(\left(\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}\right)\rightarrow\left(a,b,c\right)\Rightarrow\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=1\end{matrix}\right.\)
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 \(Q_{\max}=\frac{3}{2}\Leftrightarrow x=y=z=\sqrt{3}\)
áp dụng BĐT cô si cho 2 số ko âm
\(\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{xy}}=\dfrac{2}{\sqrt{xy}}\)
\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\)
\(\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{xz}}\)
cộng các vế vs nhau ta đc
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\)
<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\) (đpcm)
ta có : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{\sqrt{xy}}-\dfrac{1}{\sqrt{yz}}-\dfrac{1}{\sqrt{zx}}\ge0\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{2}{\sqrt{xy}}+\dfrac{1}{y}+\dfrac{1}{y}-\dfrac{2}{\sqrt{yz}}+\dfrac{1}{z}+\dfrac{1}{z}-\dfrac{2}{\sqrt{zx}}+\dfrac{1}{x}\ge0\)
\(\Leftrightarrow\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{y}}\right)^2+\left(\dfrac{1}{\sqrt{y}}-\dfrac{1}{\sqrt{z}}\right) ^2+\left(\dfrac{1}{\sqrt{z}}-\dfrac{1}{\sqrt{x}}\right)^2\ge0\forall x;y;z>0\)
\(\Rightarrow\left(đpcm\right)\)
áp dụng BĐT côsi ta có
\(\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{xy}}=\dfrac{2}{\sqrt{xy}}\)
\(\dfrac{1}{y}+\dfrac{1}{z}\ge2\sqrt{\dfrac{1}{yz}}=\dfrac{2}{\sqrt{yz}}\)
\(\dfrac{1}{z}+\dfrac{1}{x}\ge\dfrac{2}{\sqrt{xz}}\)
=> \(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)
=> đpcm