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Đặt \(\left(x;2y;3z\right)=\left(a;b;c\right)\Rightarrow a+b+c=2\)
\(S=\sqrt{\dfrac{ab}{ab+2c}}+\sqrt{\dfrac{bc}{bc+2a}}+\sqrt{\dfrac{ca}{ca+2b}}\)
\(S=\sqrt{\dfrac{ab}{ab+c\left(a+b+c\right)}}+\sqrt{\dfrac{bc}{bc+a\left(a+b+c\right)}}+\sqrt{\dfrac{ca}{ca+b\left(a+b+c\right)}}\)
\(S=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\)
\(S\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\Rightarrow x;y;z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\hept{\begin{cases}x=a\\2y=b\\3z=c\end{cases}}\left(a;b;c>0\right)\Rightarrow a+b+c=2\)
Khi đó \(S=\Sigma\sqrt{\frac{\frac{ab}{2}}{\frac{ab}{2}+c}}=\Sigma\sqrt{\frac{ab}{ab+2c}}=\Sigma\sqrt{\frac{ab}{ab+\left(a+b+c\right)c}}\)
\(=\Sigma\sqrt{\frac{ab}{ab+bc+ca+c^2}}=\Sigma\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\)
Áp dụng bđt Cô-si có
\(S\le\frac{\Sigma\left(\frac{a}{a+c}+\frac{b}{b+c}\right)}{2}=\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Đặt $(x,2y,3z)=(a,b,c)$. Khi đó bài toán trở thành:
Cho $a,b,c>0$ thỏa mãn $a+b+c=2$. Tìm GTLN của:
\(S=\sqrt{\frac{ab}{ab+2c}}+\sqrt{\frac{bc}{bc+2a}}+\sqrt{\frac{ca}{ac+2b}}\)
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Từ $a+b+c=2$ ta có:
\(S=\sqrt{\frac{ab}{ab+(a+b+c)c}}+\sqrt{\frac{bc}{bc+(a+b+c)a}}+\sqrt{\frac{ca}{ac+(a+b+c)b}}\)
\(=\sqrt{\frac{ab}{(c+a)(c+b)}}+\sqrt{\frac{bc}{(a+b)(a+c)}}+\sqrt{\frac{ca}{(b+c)(b+a)}}\)
Áp dụng BĐT Cauchy:
\(\sqrt{\frac{ab}{(c+a)(c+b)}}\leq \frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
\(\sqrt{\frac{bc}{(a+b)(a+c)}}\leq \frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)
\(\sqrt{\frac{ca}{(b+a)(b+c)}}\leq \frac{1}{2}\left(\frac{a}{b+a}+\frac{c}{b+c}\right)\)
Cộng theo vế:
\(S\leq \frac{1}{2}\left(\frac{a+c}{a+c}+\frac{b+c}{b+c}+\frac{a+b}{a+b}\right)=\frac{3}{2}\)
Vậy $S_{\max}=\frac{3}{2}$. Dấu "=" xảy ra khi $a=b=c$
hay $x=\frac{2}{3}; y=\frac{1}{3}; z=\frac{2}{9}$
bài này có trên OLM do a Dw ( incursion_03 ) giải nè.Đề tuyển sinh vào lớp 10 Dak Lak
![](https://rs.olm.vn/images/avt/0.png?1311)
cảm ơn b,b cho mình xin dạng tổng quát của bđt cosi bạn dùng ở trên được không ạ?
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bất đẳng thức AM - GM và kết hợp với giả thiết x + y + z = 3 ta có:
\(B=\sqrt{\dfrac{xy}{xy+z\left(x+y+z\right)}}+\sqrt{\dfrac{yz}{yz+x\left(x+y+z\right)}}+\sqrt{\dfrac{zx}{zx+y\left(x+y+z\right)}}\)
\(B=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}+\sqrt{\dfrac{yz}{\left(y+x\right)\left(z+x\right)}}+\sqrt{\dfrac{zx}{\left(z+y\right)\left(z+x\right)}}\le\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}+\dfrac{y}{y+x}+\dfrac{z}{z+x}+\dfrac{z}{z+y}+\dfrac{x}{z+x}\right)\)
\(B\le\dfrac{3}{2}\).
Đẳng thức xảy ra khi x = y = z = 1.
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+\sqrt{3x+yz}=x+\sqrt{x\left(x+y+z\right)+yz}=x+\sqrt{\left(x+y\right)\left(z+x\right)}\ge x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}\)
\(=x+\sqrt{xz}+\sqrt{xy}=\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{3y+zx}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế với vế ta có đpcm
a) \(\left\{{}\begin{matrix}a=x\\b=2y\\c=3z\end{matrix}\right.\Rightarrow a+b+c=2;a,b,c>0\)
\(\Rightarrow S=\sqrt{\dfrac{\dfrac{ab}{2}}{\dfrac{ab}{2}+c}}+\sqrt{\dfrac{\dfrac{bc}{2}}{\dfrac{bc}{2}+a}}+\sqrt{\dfrac{ca}{ca+2b}}\)
\(=\sqrt{\dfrac{ab}{ab+2c}}+\sqrt{\dfrac{bc}{bc+2a}}+\sqrt{\dfrac{ca}{ca+2b}}\)
Vì a,b,c>0 nên áp dụng BĐT AM-GM, ta có:
\(\sqrt{\dfrac{ab}{ab+2c}}=\sqrt{\dfrac{ab}{ab+\left(a+b+c\right)c}}=\sqrt{\dfrac{ab}{c^2+bc+ca+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\dfrac{a}{a+c}}.\sqrt{\dfrac{b}{b+c}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
\(\sqrt{\dfrac{bc}{bc+2a}}=\sqrt{\dfrac{bc}{\left(b+a\right)\left(c+a\right)}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\)
\(\sqrt{\dfrac{ca}{ca+2b}}=\sqrt{\dfrac{ca}{\left(c+b\right)\left(a+b\right)}}\le\dfrac{1}{2}\left(\dfrac{c}{b+c}+\dfrac{a}{a+b}\right)\)
\(\Rightarrow S\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)+\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)+\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi và chỉ khi: a=b=c=2/3=>\(\left(x,y,z\right)=\left\{\dfrac{2}{3};\dfrac{1}{3};\dfrac{2}{9}\right\}\)