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 \(\text{Σ}_{cyc}\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le2019.2020xyz\left(đpcm\right)\)

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

Phùng Gia Bảo             

xử hộ dấu = nè :)

dấu ''='' xảy ra <=> \(\hept{\begin{cases}x=y=z\\x+y+z=2019xyz\end{cases}\Leftrightarrow x=y=z=\frac{1}{\sqrt{673}}}\)

Nhìn qua thấy bậc của bđt là không đồng bậc nên hơi căng đấy...

Chú ý: \(2019=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{x+y+z}{xyz}\Rightarrow xyz=\frac{x+y+z}{2019}\)

\(LHS=\Sigma_{cyc}\frac{\sqrt{2019x^2+1}+1}{x}=\Sigma_{cyc}\frac{\sqrt{\frac{x}{y}+\frac{x^2}{yz}+\frac{x}{z}+1}+1}{x}\)( thay \(2019=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\))

\(=\Sigma_{cyc}\frac{\sqrt{\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)}+1}{x}=\Sigma_{cyc}\left[\sqrt{\frac{\left(\frac{x}{y}+1\right)}{x}.\frac{\left(\frac{x}{z}+1\right)}{x}}+\frac{1}{x}\right]\)

\(=\Sigma_{cyc}\sqrt{\left(\frac{1}{y}+\frac{1}{x}\right)\left(\frac{1}{z}+\frac{1}{x}\right)}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\frac{1}{2}\left[4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right]+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(=3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3\left(xy+yz+zx\right)}{xyz}=\frac{3\left(xy+yz+zx\right)}{\frac{\left(x+y+z\right)}{2019}}=\frac{6057\left(xy+yz+zx\right)}{x+y+z}\)

\(\le\frac{6057.\frac{\left(x+y+z\right)^2}{3}}{x+y+z}=2019\left(x+y+z\right)\)(đpcm)

Đẳng thức xảy ra khi \(x=y=z=\sqrt{\frac{3}{2019}}\)

P/s: Check hộ t phát:3

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)thì bài toán thành

Cho: \(ab+bc+ca=2019\)

Chứng minh:

\(\sqrt{2019+a^2}+\sqrt{2019+b^2}+\sqrt{2019+c^2}+\left(a+b+c\right)\le2019\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Ta có:

\(VT=\sqrt{ab+bc+ca+a^2}+\sqrt{ab+bc+ca+b^2}+\sqrt{ab+bc+ca+c^2}+\left(a+b+c\right)\)

\(VT=\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}+\left(a+b+c\right)\)

\(\le3\left(a+b+c\right)\)

\(VP=\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=2\left(a+b+c\right)+\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)\)

\(\ge3\left(a+b+c\right)\)

Tới đây bí :(

ban ten gi vay

â

gh

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Dễ dàng chứng minh được:

\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) với \(a,b,c>0\)(1)

Dấu bằng xảy ra \(\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

Theo đề bài, vì x, y, z > 0 nên áp dụng (1), ta có:

\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{zx}}+\frac{z^2}{z+\sqrt{xy}}\ge\)\(\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\)(2)

Vì x y, z > 0 nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:

\(x+y\ge2\sqrt{xy}\)(3)

Chứng mih tương tự, ta được;

\(y+z\ge2\sqrt{yz}\)(4);

\(z+x\ge2\sqrt{zx}\)(5)

Từ (3), (4), (5), ta được:

\(2\left(x+y+z\right)\ge2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)

\(\Leftrightarrow x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Leftrightarrow2\left(x+y+z\right)\ge x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Leftrightarrow\frac{1}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\)\(\frac{1}{2\left(x+y+z\right)}\)

\(\Leftrightarrow\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\frac{x+y+z}{2}\)

Mà theo đề bài, \(x+y+z\ge3\) nên:

\(\frac{x+y+z}{2}\ge\frac{3}{2}\)

Suy ra \(\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\frac{3}{2}\left(6\right)\)

Từ (2) và (6), ta được:

\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{zx}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{3}{2}\)(điều phải chứng minh)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}x=y=z\\x+y+z=3\end{cases}\Leftrightarrow x=y=z=1}\)

Vậy nếu x, y, z > 0 và \(x+y+z\ge3\)thì \(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{zx}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{3}{2}\)

\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{3}{2}\)

\(\frac{x^2}{x+\sqrt{yz}}+\frac{x+\sqrt{yz}}{4}\ge2\sqrt{\frac{x^2}{4}}=x\)

Tượng tự ta có : \(\frac{y^2}{y+\sqrt{xz}}+\frac{y+\sqrt{xz}}{4}\ge y\)

\(\frac{z^2}{z+\sqrt{xy}}+\frac{z+\sqrt{xy}}{4}\ge z\)

Cộng vế với vế của BĐT ta được : 

\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}+\frac{x+\sqrt{yz}}{4}+\frac{y+\sqrt{xz}}{4}+\frac{z+\sqrt{xy}}{4}\ge x+y+z\)

\(VT\ge x+y+z-\frac{x+y+z+\sqrt{yz}+\sqrt{xz}+\sqrt{xy}}{4}\)

mà \(\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\le x+y+z\)

\(VT\ge\frac{4\left(x+y+z\right)-2\left(x+y+z\right)}{4}=\frac{2\left(x+y+z\right)}{4}\)

mà \(x+y+z\ge3\)hay \(VT\ge=\frac{6}{4}=\frac{3}{2}\)

Dấu ''='' xảy ra <=> x = y = z = 1

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm 

\(\Rightarrow\hept{\begin{cases}\sqrt{xy}\le\frac{x+y}{2}\\\sqrt{yz}\le\frac{y+z}{2}\\\sqrt{xz}\le\frac{x+z}{2}\end{cases}}\)

Cộng theo từng vế 

\(\Rightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\frac{x+y}{2}+\frac{y+z}{2}+\frac{x+z}{2}\)

\(\Rightarrow1\le\frac{2\left(x+y+z\right)}{2}\)

\(\Rightarrow1\le x+y+z\)

\(\Rightarrow\frac{1}{2}\le\frac{x+y+z}{2}\left(1\right)\)

Ta có : \(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)

Áp dụng bất đẳng thức cộng mẫu số :

\(\Rightarrow A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)

\(\Rightarrow A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\ge\frac{x+y+z}{2}\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow\frac{1}{2}\le\frac{x+y+z}{2}\le\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)

\(\Rightarrow\frac{1}{2}\le\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)

Vậy GTNN của \(A=\frac{1}{2}\)

Dấu " = " xảy ra khi và chỉ khi \(x=y=z=\frac{1}{3}\)

Chúc bạn học tốt !!!

Ta có: \(1=\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\)

=> \(x+y+z\ge1\)

Có: \(A\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\ge\frac{1}{2}\)

Dấu "=" xảy ra <=> x = y = z =1/3

Vậy min A = 1/2 <=> x = y = z = 1/3

Áp dụng BĐT Cauchy cho 2 bộ số thực không âm:

\(\Rightarrow\left\{\begin{matrix}\sqrt{xy}\le\frac{x+y}{2}\\\sqrt{yz}\le\frac{y+z}{2}\\\sqrt{xz}\le\frac{x+z}{2}\end{matrix}\right.\)

Cộng theo từng vế:

\(\Rightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\frac{x+y}{2}+\frac{y+z}{2}+\frac{x+z}{2}\)

\(\Rightarrow1\le\frac{2\left(x+y+z\right)}{2}\)

\(\Rightarrow1\le x+y+z\)

\(\Rightarrow\frac{1}{2}\le\frac{x+y+z}{2}\) ( 1 )

Ta có: \(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)

Áp dụng bất đẳng thức cộng mẫu số:

\(\Rightarrow A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)

\(\Rightarrow A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\ge\frac{x+y+z}{2}\) ( 2 )

Từ điều ( 1 ) và ( 2 )

\(\Rightarrow\frac{1}{2}\le\frac{x+y+z}{2}\le\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)

\(\Rightarrow\frac{1}{2}\le\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)

Vậy GTNN của \(A=\frac{1}{2}\)

Dấu " = " xảy ra khi và chỉ khi \(x=y=z=\frac{1}{3}\)