ĐK: \(y\ne0,xy\ge0\).

\(4x^2+9y^2=16xy\)

Chia cả hai vế cho \(y^2\)ta được: 

\(4\left(\frac{x}{y}\right)^2+9=\frac{16x}{y}\)

\(\Leftrightarrow\frac{x}{y}=\frac{4\pm\sqrt{7}}{2}\)

Với \(y>0\)thì \(x\ge0\)

\(P=\frac{\sqrt{xy}+\sqrt{y^2}}{y}-\sqrt{\frac{x}{y}}=\frac{\sqrt{x}\sqrt{y}+y}{y}-\sqrt{\frac{x}{y}}=\sqrt{\frac{x}{y}}+1-\sqrt{\frac{x}{y}}=1\)

Với \(y< 0\)thì \(x\le0\):

\(P=\frac{\sqrt{xy}+\sqrt{y^2}}{y}-\sqrt{\frac{x}{y}}=\frac{\sqrt{-x}\sqrt{-y}-y}{y}-\sqrt{\frac{x}{y}}=-\sqrt{\frac{x}{y}}-1-\sqrt{\frac{x}{y}}=-2\sqrt{\frac{x}{y}}-1\)

\(=-2\sqrt{\frac{4\pm\sqrt{7}}{2}}-1=-\left(1\pm\sqrt{7}\right)-1=-2\pm\sqrt{7}\)

\(P=\frac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{y}-\frac{\sqrt{x}}{\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{y}}-\frac{\sqrt{x}}{\sqrt{y}}=\frac{\sqrt{y}}{\sqrt{y}}=1\)

ĐK : x , y > 0 

Sai rồi cha nội \(\sqrt{y}.\sqrt{y}\)=\(y\)chứ có phải \(\sqrt{y^2}\)đâu

\(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge3\left(\frac{x}{y}+\frac{y}{x}\right) \Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2+2\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\)(1)

Đặt \(t=\frac{x}{y}+\frac{y}{x}\), (1) trở thành \(t^2-3t+2\ge0\)(2)

(2) đúng khi \(t\le1\)hoặc \(t\ge2\), chú ý rằng theo bất đẳng thức AM - GM, ta có:

\(t=\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{xy}{xy}}=2\)với x,y > 0 

Do đó (2) đúng, suy ra (1) đúng ( đpcm ).

Đề đúng không thế bạn. 3 hay là 2 thế

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

Từ gt => \(\hept{\begin{cases}\left(\frac{1}{\sqrt{2}}-\sqrt{x}\right)\left(\frac{1}{\sqrt{2}}-\sqrt{y}\right)\ge0\Leftrightarrow\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}+\sqrt{2}\sqrt{xy}\left(1\right)\\x\sqrt{x}\le x\cdot\frac{1}{\sqrt{2}};y\sqrt{y}\le y\cdot\frac{1}{\sqrt{2}}\Rightarrow x\sqrt{x}+y\sqrt{y}\le\frac{1}{\sqrt{2}}\left(x+y\right)\left(2\right)\end{cases}}\)

Lại có \(\hept{\begin{cases}\sqrt{xy}\le xy+\frac{1}{4}\\\sqrt{xy}\le\frac{x+y}{2}\end{cases}\Rightarrow\hept{\begin{cases}\frac{2\sqrt{2}}{3}\sqrt{xy}\le\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)\left(3\right)\\\frac{\sqrt{2}}{3}\sqrt{xy}\le\frac{\sqrt{2}}{6}\left(x+y\right)\left(4\right)\end{cases}}}\)

Từ (1)(2)(3)(4) ta có:\(x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}\left(x+y\right)+\frac{\sqrt{2}}{2}+\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)+\frac{\sqrt{2}}{6}\left(x+y\right)\)

\(\le\frac{2\sqrt{2}}{3}\left(1+x+y+xy\right)\)

=> \(VT=\frac{\sqrt{x}}{1+y}+\frac{\sqrt{y}}{1+x}=\frac{x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}}{1+x+y+xy}\le\frac{2\sqrt{2}}{3}\)

Dấu "=" xảy ra <=> x=y=\(\frac{1}{2}\)

@Lightning Farron

Dễ thấy:

     \(VT\ge\left(x+y\right)^2+1-\dfrac{\left(x+y\right)^2}{4}=\dfrac{3\left(x+y\right)^2}{4}+1\)

Áp dụng Cô-si:

     \(\dfrac{3\left(x+y\right)^2}{4}+1\ge2\sqrt{\dfrac{3\left(x+y\right)^2}{4}.1}=\sqrt{3}\left|x+y\right|\ge\sqrt{3}\left(x+y\right)\)

Do đó:

     \(\left(x+y\right)^2+1-xy\ge\sqrt{3}\left(x+y\right),\forall x,y\in R\)