Ta có với x,y,z >0 thì:\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)
Bất đẳng thức Cô si ta có:
\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{x\sqrt{1-x^2}}\ge2\\ \Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\Leftrightarrow\dfrac{x^2}{\sqrt{1-x^2}}\ge2x^3\)
Tương tự: \(\dfrac{y^2}{\sqrt{1-y^2}}\ge2y^3;\dfrac{z^2}{\sqrt{1-z^2}}\ge2z^3\)
Từ đó ta có:\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\left(dpcm\right)\)
 

em không biết

gà quá

Từ giả thiết \(x+y+z=xyz=\frac{1}{xy}\)\(=\frac{1}{yz}\)\(=\frac{1}{zx}\)\(=1\)

Đặt \(\frac{1}{x}\)\(=a,\frac{1}{y}\)\(=b,\frac{1}{z}\)\(=c=ab+bc+ca=1\)

Ta có :

\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\)\(=\sqrt{\frac{1}{\sqrt{1+x^2}}+}+\sqrt{\frac{1}{\sqrt{1+y^2}}+\sqrt{\frac{1}{\sqrt{1+z^2}}}}\)

\(=\sqrt{\frac{1}{x}+x}+\sqrt{\frac{1}{y}+y}+\sqrt{\frac{1}{z}+z}=\sqrt{\frac{a}{a+\frac{1}{a}}}+\sqrt{\frac{b}{b+\frac{1}{b}}}\)\(+\sqrt{\frac{c}{c+\frac{1}{c}}}\)

\(=\frac{a}{\sqrt{a^2}+1}\)\(+\frac{b}{\sqrt{b^2}+1}\)\(+\frac{c}{\sqrt{c^2}+1}\)

Đến đây :

\(\frac{a}{\sqrt{a^2}+1}\)\(=\frac{a}{\left(a^2+ab+bc+ca\right)}\)\(=\frac{a}{\sqrt{\left(a+b\right)}\left(a+c\right)}\)

\(=\sqrt{\frac{a}{a+b}}\)\(\cdot\frac{a}{a+c}\)\(< \frac{1}{2}\)\(\left(\frac{b}{b+a}+\frac{b}{b+c}\right);\frac{c}{\sqrt{c^2}+1}\)\(< \frac{1}{2}\)\(\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\)

ộng 3 bất đẳng thức lại ta có điều phải chứng minh

Á nhầm nhaaa cái cuối cùng là cộng z2 đó

Ta có :

\(\frac{1+\sqrt{1+x^2}}{x}=\frac{2+\sqrt{4\left(1+x^2\right)}}{2x}\le\frac{2+\frac{4+1+x^2}{2}}{2x}=\frac{9+x^2}{4x}\)

tương tự : \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{9+y^2}{4y}\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{9+z^2}{4z}\)

\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le\frac{\left(9+x^2\right)yz+\left(9+y^2\right)xz+\left(9+z^2\right)xy}{4xyz}\)

\(=\frac{9\left(xy+yz+xz\right)+xyz\left(x+y+z\right)}{4xyz}\le\frac{9\frac{\left(x+y+z\right)^2}{3}+\left(xyz\right)^2}{4xyz}=\frac{4\left(xyz\right)^2}{4xyz}=xyz\)

Dấu " = " xảy ra khi x = y = z = \(\sqrt{3}\)

Ta có: \(\frac{1+\sqrt{1+x^2}}{x}=\frac{1+\sqrt{1\times\left(1+x^2\right)}}{x}\le\frac{1+\frac{1+1+x^2}{2}}{x}=\frac{2+\frac{x^2}{2}}{x}=\frac{2}{x}+\frac{x}{2}\)(Áp dụng bđt Cauchy ở chỗ \(\sqrt{1\times\left(1+x^2\right)}\)

Tương tự với b,c . Ta được VT\(\le\)\(\frac{x}{2}+\frac{y}{2}+\frac{z}{2}+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\)

\(\le\)\(\frac{x+y+z}{2}\)+ \(\frac{2xy+2yz+2xz}{xyz}\)= \(\frac{x+y+z}{2}\)+ \(\frac{4xy+4yz+4xz}{2xyz}\)= \(\frac{xyz}{2}+\frac{4xy+4yz+4xz}{2xyz}\)

Ta chứng minh được \(4xy+4yz+4xz\le\left(x+y+z\right)^2\)bằng phương pháp biến đổi tương đương

=> VT \(\le\)\(\frac{xyz}{2}+\frac{\left(x+y+z\right)^2}{2xyz}=\frac{xyz}{2}+\frac{\left(xyz\right)^2}{2xyz}=\frac{xyz}{2}+\frac{xyz}{2}=xyz\)(Điều phải cm)

Dấu = xảy ra <=> 

Sử dụng BĐT AM-GM, ta có: 

\(x^3+y^2\ge2yx\sqrt{x}\)

\(\Rightarrow\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2yx\sqrt{x}}=\frac{1}{xy}\)

Tương tự cộng lại suy ra: 

\(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Theo AM-GM: \(x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\)

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

Tương tự: \(\frac{2\sqrt{y}}{y^3+z^2}\le\frac{1}{yz}\)

\(\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{zx}\)

Cộng vế với vế => \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)

Theo AM-GM; \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}}{2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Dấu " = " xảy ra <=> x=y=z=1

Áp dụng bất đẳng thức Cacuhy - Schwarz 

\(\Rightarrow\hept{\begin{cases}x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\\y^3+z^2\ge2\sqrt{y^3z^2}=2yz\sqrt{y}\\z^3+x^2\ge2\sqrt{z^3x^2}=2xz\sqrt{z}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\\\frac{2\sqrt{y}}{y^3+z^2}\le\frac{2\sqrt{y}}{2yz\sqrt{y}}=\frac{1}{yz}\\\frac{2\sqrt{z}}{z^3+x^2}\le\frac{2\sqrt{z}}{2xz\sqrt{z}}=\frac{1}{xz}\end{cases}}\)

\(\Rightarrow VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(1\right)\)

Áp dụng bất đẳng thức Cacuchy Schwarz 

\(\Rightarrow\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2y^2}}=\frac{2}{xy}\\\frac{1}{y^2}+\frac{1}{z^2}\ge2\sqrt{\frac{1}{y^2z^2}}=\frac{2}{yz}\\\frac{1}{z^2}+\frac{1}{x^2}\ge2\sqrt{\frac{1}{z^2x^2}}=\frac{2}{xz}\end{cases}}\)

\(\Rightarrow2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\ge2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(2\right)\)

Từ (1) và (2)

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

\(\Leftrightarrow\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\left(đpcm\right)\)

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

\(=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\) (áp dụng BĐT quen thuộc \(ab+bc+ca\le a^2+b^2+c^2\))

Đẳng thức xảy ra khi x = y = z = 1

Sửa đề : \(\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\hept{\begin{cases}x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\\y^3+z^2\ge2\sqrt{y^3z^2}=2yz\sqrt{y}\\z^3+x^2\ge2\sqrt{z^3x^2}=2xz\sqrt{z}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\\\frac{2\sqrt{y}}{y^3+z^2}\le\frac{2\sqrt{y}}{2yz\sqrt{y}}=\frac{1}{yz}\\\frac{2\sqrt{z}}{z^3+x^2}\le\frac{2\sqrt{z}}{2xz\sqrt{z}}=\frac{1}{xz}\end{cases}}\)

\(\Rightarrow VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(1\right)\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2y^2}}=\frac{2}{xy}\\\frac{1}{y^2}+\frac{1}{z^2}\ge2\sqrt{\frac{1}{y^2z^2}}=\frac{2}{yz}\\\frac{1}{z^2}+\frac{1}{x^2}\ge2\sqrt{\frac{1}{x^2z^2}}=\frac{2}{xz}\end{cases}}\)

\(\Rightarrow2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\ge2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(2\right)\)

Từ (1) và (2) :

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

\(\Leftrightarrow\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\left(đpcm\right)\)

Chúc bạn học tốt !!!

ĐỊT MẸ

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

Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow ab+bc+ca=1\)

\(P=\frac{a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}=\frac{a}{\sqrt{ab+bc+ca+a^2}}+\frac{b}{\sqrt{ab+bc+ca+b^2}}+\frac{c}{\sqrt{ab+bc+ca+c^2}}\)

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

\(P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{c}{b+c}\right)=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\) hay \(x=y=z=\sqrt{3}\)

Thay giá trị x = y = z vô thì thấy VT > 2 nên nghi ngờ đề sai. B xem lại