Cho x + y + z = 1. Tìm GTLN của:
A = 4xy + yz + 5zx
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MD
1
16 tháng 7 2021
\(=>A=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)
áp dụng BĐT AM-GM
\(=>\sqrt{x-1}\le\dfrac{x-1+1}{2}=\dfrac{x}{2}\)
\(=>\dfrac{\sqrt{x-1}}{x}\le\dfrac{\dfrac{x}{2}}{x}=\dfrac{1}{2}\left(1\right)\)
có \(\dfrac{\sqrt{y-2}}{y}=\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\)
\(=>\sqrt{\left(y-2\right)2}\le\dfrac{y-2+2}{2}=\dfrac{y}{2}\)
\(=>\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\le\dfrac{\dfrac{y}{2}}{\sqrt{2}.y}=\dfrac{1}{2\sqrt{2}}\left(2\right)\)
tương tự \(=>\dfrac{\sqrt{z-3}}{z}\le\dfrac{1}{2\sqrt{3}}\left(3\right)\)
(1)(2)(3)\(=>A\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)
HT
0
N
3
G
14 tháng 5 2019
Ta có: \(x+y+z=1\Rightarrow\hept{\begin{cases}\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\\\sqrt{y+xz}=\sqrt{y\left(x+y+z\right)+xz}=\sqrt{\left(x+y\right)\left(y+z\right)}\\\sqrt{z+xy}=\sqrt{z\left(x+y+z\right)+xy}=\sqrt{\left(x+z\right)\left(y+z\right)}\end{cases}}\)
Ta viết lại A
\(A=\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(y+z\right)\left(x+z\right)}\)
Áp dụng bđt AM-GM:
\(A\le\frac{x+y+x+z+x+y+y+z+y+z+x+z}{2}=2\)
\("="\Leftrightarrow x=y=z=\frac{1}{3}\)
14 tháng 5 2019
\(x+yz=x\left(x+y+z\right)+yz\)
\(=x^2+xy+xz+yz\)
\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)
+ Tương tự : \(y+xz=\left(x+y\right)\left(y+z\right)\)
\(z+xy=\left(x+z\right)\left(y+z\right)\)
+ Theo bđt AM-GM : \(\sqrt{\left(x+y\right)\left(x+z\right)}\le\frac{x+y+x+z}{2}\)
\(\Rightarrow\sqrt{\left(x-1\right)\left(y-1\right)}\le\frac{2x+y+z}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x+y=x+z\Leftrightarrow y=z\)
+ Tương tự ta cm đc :
\(\sqrt{\left(x+y\right)\left(y+z\right)}\le\frac{x+2y+z}{2}\). Dấu "=" xảy ra \(\Leftrightarrow x=z\)
\(\sqrt{\left(x+z\right)\left(y+z\right)}\le\frac{x+y+2z}{2}\). Dấu "=" xảy ra \(\Leftrightarrow x=y\)
Do đó : \(A\le\frac{4\left(x+y+z\right)}{2}=2\)
A = 2 \(\Leftrightarrow x=y=z=\frac{1}{3}\)
Vậy Max A = 2 \(\Leftrightarrow x=y=z=\frac{1}{3}\)
NT
1
3 tháng 12 2021
\(x^2+y^2+z^2+4xyz=2\left(xy+yz+zx\right)\\ \Leftrightarrow\left(x-y-z\right)^2=\left(1-x\right)4yz\ge0\\ \Leftrightarrow1-x\ge0\Leftrightarrow0< x\le1\\ \Leftrightarrow\left(x-y-z\right)^2=\left(1-x\right)4yz\le\left(1-x\right)\left(y+z\right)^2\\ \Leftrightarrow x^2-2x\left(y+z\right)+\left(y+z\right)^2\le\left(1-x\right)\left(y+z\right)^2\\ \Leftrightarrow x^2-2x\left(y+z\right)\le\left(y+z\right)^2\left(1-x-1\right)=-x\left(y+z\right)^2\\ \Leftrightarrow x-2\left(y+z\right)\le-\left(y+z\right)^2\\ \Leftrightarrow x\le\left(y+z\right)\left[2-\left(y+z\right)\right]\)
Đặt \(2-\left(y+z\right)=t\)
\(P=x\left(1-y\right)\left(1-z\right)\le x\left(\dfrac{1-y+1-z}{2}\right)^2=\dfrac{x\left[2-\left(y+z\right)\right]^2}{4}\\ \Leftrightarrow4P\le x\left[2-\left(y+z\right)\right]^2\le\left(y+z\right)\left[2-\left(y+z\right)\right]^3\\ \Leftrightarrow4P\le t^3\left(2-t\right)=\dfrac{27}{16}-\dfrac{\left(4t^2+4t+3\right)\left(2t-3\right)^2}{16}\)
Mà \(-\dfrac{\left(4t^2+4t+3\right)\left(2t-3\right)^2}{16}\le0\Leftrightarrow4P\le\dfrac{27}{16}\Leftrightarrow P\le\dfrac{27}{64}\)
Dấu \("="\Leftrightarrow x=\dfrac{3}{4};y=z=\dfrac{1}{4}\)
\(A-\frac{5}{4}\left(x+y+z\right)^2=-\frac{1}{20}\left(3y+5z-5x\right)^2-\frac{4y^2}{5}\le0\)
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