Tìm x, y,z nguyên dương sao cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)
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12 tháng 3 2021
\(\dfrac{x}{x^2+yz}+\dfrac{y}{y^2+zx}+\dfrac{z}{z^2+xy}\le\dfrac{x}{2\sqrt{x^2yz}}+\dfrac{y}{2\sqrt{y^2zx}}+\dfrac{z}{2\sqrt{z^2xy}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}+\dfrac{1}{\sqrt{xy}}\right)\le\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\).
Đẳng thức xảy ra khi x = y = z = 1.
8 tháng 8 2021
chắc đề cho x+y+z=1
\(=>\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\)
\(=>\dfrac{x}{x+\sqrt{\left(x+y\right)\left(x+z\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}\)
\(=\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
làm tương tự với \(\dfrac{y}{y+\sqrt{y+xz}},\dfrac{z}{z+\sqrt{z+xy}}\)
\(=>A\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) dấu"=" xảy ra<=>x=y=z=`/3
TT
7 tháng 1 2021
b/ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\dfrac{a}{d}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
=> \(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{c+d+b}\right)^3\) (2)Từ (1) và (2)=>đpcm
9 tháng 12 2021
\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)
\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)
9 tháng 12 2021
\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
26 tháng 1 2018
+,3 = x + y + z \(\ge\) \(3\sqrt[3]{xyz}\Rightarrow xyz\le1\)
+, P \(\ge\) \(3\sqrt[3]{\dfrac{1}{xyz\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\ge\dfrac{3}{\sqrt[3]{\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\ge\dfrac{3}{\dfrac{x+y+z+3}{3}}=\dfrac{3}{2}\)
Không mất tính tổng quát, giả sử \(a\le b\le c\)
\(\Rightarrow\dfrac{1}{a}\ge\dfrac{1}{b}\ge\dfrac{1}{c}\)
\(\Rightarrow2=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}=\dfrac{3}{a}\)
\(\Rightarrow a\le\dfrac{3}{2}\)
Mà a là số nguyên dương
\(\Rightarrow a=1\)
Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)
\(\Rightarrow\dfrac{1}{b}+\dfrac{1}{c}=1\le\dfrac{1}{b}+\dfrac{1}{b}=\dfrac{2}{b}\)
\(\Rightarrow b\le2\)
\(\Rightarrow y\in\left\{1;2\right\}\)
\(\Rightarrow z\in\left\{1;2\right\}\)
Vậy \(\left(x;y;z\right)\in\left\{\left(1;2;2\right),\left(2;2;1\right),\left(2;1;2\right),\left(2;2;1\right)\right\}\)