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đặt \(\dfrac{x+2y}{3}=\dfrac{y+2z}{4}=\dfrac{z+2x}{5}=t\)
vậy ta đc \(\left\{{}\begin{matrix}x+2y=3t\left(1\right)\\y+2z=4t\left(2\right)\\z+2x=5t\left(3\right)\end{matrix}\right.\)
từ (1) ta có: x = 3t - 2y
thay vào (3) ta được: z + 2 × (3t - 2y) = 5t
=> z + 6t - 4y = 5t => z = -t + 4y (3')
từ (2) ta có: \(z=\dfrac{4t-y}{2}\left(2'\right)\)
từ (2') và (3') ta có:
\(-t+4y=\dfrac{4t-y}{2}\\ -2t+8y=4t-y\\ 9y=6t=>y=\dfrac{2}{3}t\)
thay vào (1): \(x=3t-2\cdot\dfrac{2}{3}t=3t-\dfrac{4}{3}t=\dfrac{5}{3}t\)
thay vào (2'): \(z=\dfrac{4t-\dfrac{2}{3}t}{2}=\dfrac{\dfrac{10}{3}t}{2}=\dfrac{5}{3}t\)
vậy: \(x=\dfrac{5}{3}t;y=\dfrac{2}{3}t;z=\dfrac{5}{3}t\)
thay các giá trị này vào biểu thức trên ta được:
\(xy+yz+2zx=\dfrac{5}{3}t\cdot\dfrac{2}{3}t+\dfrac{2}{3}t\cdot\dfrac{5}{3}t+\dfrac{5}{3}t\cdot\dfrac{5}{3}t\\ xy+yz+2zx=\dfrac{10}{9}t^2+\dfrac{10}{9}t^2+\dfrac{50}{9}t^2\\ =>\dfrac{70}{9}t^2=280=>t=6\\ \left\{{}\begin{matrix}x=\dfrac{5}{3}t=\dfrac{5}{3}\cdot6=10\\y=\dfrac{2}{3}t=\dfrac{2}{3}\cdot6=4\\y=\dfrac{5}{3}t=\dfrac{5}{3}\cdot6=10\end{matrix}\right.\)
vậy các số x; y; z cần tìm lần lượt là 10; 4; 10
Đặt \(\hept{\begin{cases}2x+y+z=4a\\2y+x+z=4b\\2z+x+y=4c\end{cases}\Rightarrow}\hept{\begin{cases}x=3a-b-c\\y=3b-c-a\\z=3c-a-b\end{cases}}\)thay vào biểu thức đó
\(\Rightarrow\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\)
\(=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}\)
\(=\frac{3}{4}-\frac{b-c}{4a}+\frac{3}{4}-\frac{c-a}{4b}+\frac{3}{4}-\frac{a-b}{4c}\)
\(=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\)
Áp dụng BĐT sau: \(\frac{a}{b}+\frac{b}{a}\ge2\Rightarrow\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\ge6\)
\(\Leftrightarrow\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\ge\frac{6}{4}\)
\(\Leftrightarrow\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\le\frac{3}{4}\)
Từ đó ta có: \(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\le\frac{3}{4}\)(đpcm).
Dấu "=" xảy ra <=> x=y=z.
Theo Cauchy Schwarz:
\(\frac{x}{2x+y+z}=\frac{x}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
Tương tự:
\(\frac{y}{2y+z+x}\le\frac{1}{4}\left(\frac{y}{y+x}+\frac{y}{y+z}\right);\frac{z}{2z+y+x}\le\frac{1}{4}\left(\frac{z}{z+y}+\frac{z}{z+x}\right)\)
Cộng lại:
\(D\le\frac{3}{4}\left(đpcm\right)\)
Áp dụng bđt Cauchy-Schwarz:
\(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\)
\(=\frac{x}{\left(x+y\right)+\left(x+z\right)}+\frac{y}{\left(x+y\right)+\left(y+z\right)}+\frac{z}{\left(y+z\right)+\left(x+z\right)}\)
\(\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{x+y}+\frac{y}{y+z}+\frac{z}{y+z}+\frac{z}{x+z}\right)=\frac{3}{4}\)
\("="\Leftrightarrow x=y=z\)
Cậu vào đây nha !
Câu hỏi của doanthihuong - Toán lớp 7 - Học toán với OnlineMath
Đặt \(a=2x+y+z;b=2y+z+x;c=2z+x+y\)
\( \implies\) \(a+b+c=\left(2x+y+z\right)+\left(2y+z+x\right)+\left(2z+x+y\right)\)
\( \implies\) \(a+b+c=4x+4y+4z\)
\( \implies\) \(x+y+z=\frac{a+b+c}{4}\)
+)Ta có : \(a=2x+y+z\)
\(\iff\) \(a=x+\left(x+y+z\right)\)
\(\iff\) \(a-\left(x+y+z\right)=x\)
\(\iff\) \(a-\frac{a+b+c}{4}=x\)
\(\iff\) \(x=\frac{3a-b-c}{4}\)
+)Ta có :\(b=2y+z+x\)
\(\iff\) \(b=y+\left(y+z+x\right)\)
\(\iff\)\(b-\left(y+z+x\right)=y\)
\(\iff\) \(b-\frac{a+b+c}{4}=y\)
\(\iff\)\(y=\frac{3b-c-a}{4}\)
+)Ta có :\(c=2z+x+y\)
\(\iff\) \(c=z+\left(z+x+y\right)\)
\(\iff\) \(c-\left(z+x+y\right)=z\)
\(\iff\) \(c-\frac{a+b+c}{4}=z\)
\(\iff\)\(z=\frac{3c-a-b}{4}\)
\( \implies\) \(\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\)
\(=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}\)
\(=\frac{9}{4}-\left(\frac{b}{4a}+\frac{c}{4a}+\frac{c}{4b}+\frac{a}{4b}+\frac{a}{4c}+\frac{b}{4c}\right)\)
\(=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\)
\(=\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\)
Áp dụng bất đẳng thức ( BĐT Cosi ) : \(m+n\)\( \geq\)\(2\sqrt{mn}\) \(\left(m;n>0\right)\)ta được :
\(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{a}.\frac{a}{b}}\) = 2 \( \implies\) \(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2
\(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2 \(\sqrt{\frac{c}{a}.\frac{a}{c}}\) = 2 \( \implies\) \(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2
\(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{c}.\frac{c}{b}}\) = 2 \( \implies\) \(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2
\( \implies\) \(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\) \( \geq\) 2 + 2 + 2
\( \implies\) \(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\) \( \geq\) 6
\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{6}{4}\)
\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{3}{2}\)
\( \implies\) \(-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(-\frac{3}{2}\)
\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{9}{4}-\frac{3}{2}\)
\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{3}{4}\)