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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)
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 \(x=y=z=3\)
Áp dụng BĐT AM-GM:
\(VT=\sum\dfrac{\sqrt{\left(x+y\right)^2-xy}}{4yz+1}\ge\sum\dfrac{\sqrt{\left(x+y\right)^2-\dfrac{1}{4}\left(x+y\right)^2}}{\left(y+z\right)^2+1}=\sum\dfrac{\dfrac{\sqrt{3}}{2}\left(x+y\right)}{\left(y+z\right)^2+1}\)
Set \(\left\{{}\begin{matrix}x+y=a\\y+z=b\\z+x=c\end{matrix}\right.\)thì giả thiết trở thành \(a+b+c=3\) và cần chứng minh \(\dfrac{\sqrt{3}}{2}.\sum\dfrac{a}{b^2+1}\ge\dfrac{3\sqrt{3}}{4}\)
\(\Leftrightarrow\sum\dfrac{a}{b^2+1}\ge\dfrac{3}{2}\)( đến đây quen thuộc rồi)
Ta có:\(\sum\dfrac{a}{b^2+1}=\sum a-\sum\dfrac{ab^2}{b^2+1}\ge3-\sum\dfrac{ab^2}{2b}\)(AM-GM)
\(VT\ge3-\sum\dfrac{ab}{2}\ge3-\dfrac{\dfrac{1}{3}\left(a+b+c\right)^2}{2}=\dfrac{3}{2}\)( AM-GM)
Vậy ta có đpcm.Dấu = xảy ra khi a=b=c=1 hay \(x=y=z=\dfrac{1}{2}\)
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)
Lời giải:
Đặt \((\sqrt{x}, \sqrt{y}, \sqrt{z})=(a,b,c)\Rightarrow abc=1\)
Bài toán trở thành chứng minh:
\(\frac{1}{(ab+a+1)^2}+\frac{1}{(bc+b+1)^2}+\frac{1}{(ca+c+1)^2}\geq \frac{1}{3}\)
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Áp dụng 1 kết quả quen thuộc của BĐT AM-GM: \(x^2+y^2+z^2\geq \frac{(x+y+z)^2}{3}\) ta có:
\(\frac{1}{(ab+a+1)^2}+\frac{1}{(bc+b+1)^2}+\frac{1}{(ca+c+1)^2}\geq \frac{1}{3}\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)^2\)
Mà:
\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}=\frac{c}{abc+ac+c}+\frac{ac}{bc.ac+b.ac+ac}+\frac{1}{ac+c+1}\)
\(=\frac{c}{1+ac+c}+\frac{ac}{c+1+ac}+\frac{1}{ac+c+1}=\frac{ac+c+1}{ac+c+1}=1\) (thay $abc=1$)
Do đó:
\(\frac{1}{(ab+a+1)^2}+\frac{1}{(bc+b+1)^2}+\frac{1}{(ca+c+1)^2}\geq \frac{1}{3}.1^2=\frac{1}{3}\) (đpcm)
Dâu bằng xảy ra khi $a=b=c=1$ hay $x=y=z=1$
Lời giải:
Ta có:
\(\frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\)
\(=\frac{x}{\sqrt{x^2+xy+yz+xz}}+\frac{y}{\sqrt{y^2+xy+yz+xz}}+\frac{z}{\sqrt{z^2+xy+yz+xz}}\)
\(=\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(y+z)(y+x)}}+\frac{z}{\sqrt{(z+x)(z+y)}}\)
Áp dụng BĐT Cauchy:
\(\frac{x}{\sqrt{(x+y)(x+z)}}\leq \frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
\(\frac{y}{\sqrt{(y+z)(y+x)}}\leq \frac{1}{2}\left(\frac{y}{y+z}+\frac{y}{y+x}\right)\)
\(\frac{z}{\sqrt{(z+x)(z+y)}}\leq \frac{1}{2}\left(\frac{z}{z+x}+\frac{z}{z+y}\right)\)
Cộng theo vế:
\(\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(y+z)(y+x)}}+\frac{z}{\sqrt{(z+x)(z+y)}}\leq \frac{1}{2}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{3}{2}\)
Ta có đpcm
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
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 $x=y=z=3$
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\)
\(\Rightarrow\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}\ge\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\)
\(\Rightarrow\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}-\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\ge0\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{2}{\sqrt{xy}}+\dfrac{1}{y}+\dfrac{1}{y}-\dfrac{2}{\sqrt{yz}}+\dfrac{1}{z}+\dfrac{1}{z}-\dfrac{2}{\sqrt{zx}}+\dfrac{1}{x}\ge0\)
\(\Rightarrow\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{y}}\right)^2+\left(\dfrac{1}{\sqrt{y}}-\dfrac{1}{\sqrt{z}}\right)^2+\left(\dfrac{1}{\sqrt{z}}-\dfrac{1}{\sqrt{x}}\right)^2\ge0\) (luôn đúng)
Dấu = xảy ra khi \(x=y=z\)