Khi \(\frac{x+y+z}{3}=\sqrt{673}.\)Chứng minh \(xy+yz+zx\text{≤}2019\)
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17 tháng 2 2019
Đk: $x\geq \frac{1}{2}$
Pt $\Leftrightarrow 4x^2+3x-7=4(\sqrt{x^3+3x^2}-2)+2(\sqrt{2x-1}-1)$
$\Leftrightarrow +4\frac{(x-1)(x+2)^2}{\sqrt{x^3+3x^2}+2}+4\frac{x-1}{\sqrt{2x-1}+1}-(x-1)(4x+7)=0$
$\Leftrightarrow (x-1)[\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-(4x+7)]=0$
$\Leftrightarrow x=1\vee \frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7=0$ $(*)$
Xét hàm số $f(x)=\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7,x\in [\frac{1}{2};+\infty )$ thì $f(x)>0,\forall x\in [\frac{1}{2};+\infty )$
$\Rightarrow $ Pt $(*)$ vô nghiệm
TN
26 tháng 2 2018
\(VT=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\)
\(\ge\frac{3x}{y+z+1}+\frac{3y}{x+z+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+xz+x}+\frac{3y^2}{xy+yz+y}+\frac{3z^2}{xz+yz+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x^2+y^2+z^2}\)
\(\ge\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\ge xy+yz+xz=VP\)
Dấu "=" <=> x=y=z=1
TN
17 tháng 1 2017
Bài 1:Áp dụng C-S dạng engel
\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)
NT
0
T
20 tháng 10 2020
1111111111111111111
\(VT=\Sigma\frac{xy+yz+zx}{xy}=3+\Sigma\frac{z\left(x+y\right)}{xy}\)
Đến đây để ý \(\frac{1}{2}\left[\frac{z\left(x+y\right)}{xy}+\frac{y\left(z+x\right)}{zx}\right]\ge\sqrt{\frac{\left(z+x\right)\left(x+y\right)}{x^2}}\left(\text{AM - GM}\right)\)
Là xong.
16 tháng 8 2021
Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)
Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)
\(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
Cộng vế với vế ta có:
\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)
\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)
KN
18 tháng 2 2020
Ta có: \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=2019\)
\(\Rightarrow\frac{x+y+z}{xyz}=2019\)
\(\Rightarrow x+y+z=2019xyz\)
\(\Rightarrow2019x^2=\frac{x^2+xy+xz}{yz}\)
\(\Rightarrow2019x^2+1=\frac{x^2+xy+xz+yz}{yz}=\frac{\left(x+y\right)\left(x+z\right)}{yz}\)
\(=\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)\)
\(\Rightarrow\sqrt{2019x^2+1}=\sqrt{\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)}\)\(\le\frac{1}{2}\left(\frac{x}{y}+\frac{x}{z}+2\right)=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)(cô -si)
\(\Rightarrow\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le\frac{x^2+1+1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}\)\(=x+\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự ta có: \(\frac{y^2+1+\sqrt{2019y^2+1}}{y}\le y+\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\)
và \(\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le z+\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Cộng từng vế của các bđt trên, ta được:
\(\text{Σ}_{cyc}\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le x+y+z+3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Chứng minh được: \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3\left(xy+yz+zx\right)}{xyz}=\frac{2019.3\left(xy+yz+zx\right)}{2019xyz}\)
\(\le\frac{2019\left(x+y+z\right)^2}{x+y+z}=2019\left(x+y+z\right)\)
\(\Rightarrow VT\le2020\left(x+y+z\right)=2020.2019xyz\)
Vậy \(\text{Σ}_{cyc}\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le2019.2020xyz\left(đpcm\right)\)
21 tháng 3 2020
Theo bài ra ta có:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z}{xyz}+\frac{x}{xyz}+\frac{y}{xyz}=\frac{x+y+z}{xyz}=2019\)
\(\Rightarrow x+y+z=2019xyz\)
\(\Rightarrow2019x^2=\frac{x^2+xy+xz}{yz}\)
\(\Rightarrow2019x^2+1=\frac{x^2+xy+xz+yz}{yz}=\frac{\left(x+y\right)\left(x+z\right)}{yz}=\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)\)
\(\Rightarrow\sqrt{2019x^2+1}=\sqrt{\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)}\le\frac{1}{2}\left(\frac{x}{y}+\frac{x}{z}+2\right)=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)(Theo BĐT Cosi)
\(\Rightarrow\frac{x^2+1+\sqrt{2019^2+1}}{x}\le\frac{x+1+1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=x+\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự:
\(\frac{y^2+1+\sqrt{2019y^2+1}}{y}\le y+\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\)
\(\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le z+\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(\Rightarrow VT\le x+y+z+3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Chứng minh được: \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3\left(xy+yz+zx\right)}{xyz}=\frac{2019\cdot3\left(xy+yz+zx\right)}{2019xyz}\le\frac{2019\left(x+y+z\right)^2}{x+y+z}\)\(=2019\left(x+y+z\right)\)
\(\Rightarrow VT\le2020\left(x+y+z\right)=2020\cdot2019xyz=VP\)
=> ĐPCM
\(\frac{x+y+z}{3}=\sqrt{673}\). Bình phương hai vế \(\Rightarrow\left(\frac{x+y+z}{3}\right)^2=\left(\sqrt{673}\right)^2\)
\(\Leftrightarrow\frac{\left(x+y+z\right)^2}{9}=673\Leftrightarrow\frac{x^2+y^2+z^2+2\left(xy+xz+yz\right)}{3}=673.3=2019\)
Tiếp theo bạn chứng minh \(x^2+y^2+z^2\ge xy+xz+yz\)
( Dễ thôi, nhân đôi hai vế rồi chuyến sang vế trái tách ghép là được 3 hằng đẳng thức luôn \(\ge0\) )
Sau khi chứng minh được thì tiếp tục cái đẳng thức trên : v
\(\Rightarrow\frac{x^2+y^2+z^2+2\left(xy+xz+yz\right)}{3}\le\frac{xy+xz+yz+2\left(xy+xz+yz\right)}{3}\)
\(=\frac{3\left(xy+xz+yz\right)}{3}=xy+xz+yz\Rightarrow\frac{\left(x+y+z\right)^2}{3}\ge xy+xz+yz\)
. Vì \(\frac{x^2+y^2+z^2+2\left(xy+xz+yz\right)}{3}=\frac{\left(x+y+z\right)^2}{3}=2019\)
\(\Rightarrow\frac{\left(x+y+z\right)^2}{3}\ge xy+xz+yz\Leftrightarrow xy+xz+yz\le2019\) ( đpcm )