Ta có: \(2\left(a^5+b^5\right)=\left(a+b\right)\left(a^5+b^5\right)\ge\left(a^3+b^3\right)^2\)

\(\Rightarrow a^5+b^5\ge\frac{\left(a^3+b^3\right)^2}{2}\)

Mà \(2\left(a^3+b^3\right)=\left(a+b\right)\left(a^3+b^3\right)\ge\left(a^2+b^2\right)^2\)

\(\Rightarrow a^5+b^5\ge\frac{\left(\frac{\left(a^2+b^2\right)^2}{2}\right)^2}{2}=\frac{\left(a^2+b^2\right)^4}{8}\)

\(\ge\frac{\left(\frac{\left(a+b\right)^2}{2}\right)^4}{8}=\frac{16}{8}=2\left(đpcm\right)\)

Đặt \(x=1+a\) \(\Rightarrow y=1-a\)

\(\Rightarrow x^5+y^5=\left(1+a\right)^5+\left(1-a\right)^5\)

\(=10a^4+20a^2+2\ge2\) ( vì \(a^4>0;a^2>0\) với mọi a )

\(\Rightarrow x^5+y^5\ge2\left(ĐPCM\right)\)

Dấu = xảy ra khi \(a=0\Leftrightarrow x=y=1\)

- Bửa hổm mới làm nà ~~

- Mà làm sao quên ròi

\(\left(x+y\right)^2+\frac{x+y}{2}=\left(x+y\right)\left(x+\frac{1}{4}+y+\frac{1}{4}\right)\ge2\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\ge2x\sqrt{y}+2y\sqrt{x}\)

Dau '=' xay ra khi \(x=y=\frac{1}{4}\)

Theo giả thiết \(x-y>0\). Do đó theo bất đẳng thức Cô-Si ta có

\(\frac{x^2+y^2}{x-y}=\frac{\left(x-y\right)^2+2xy}{x-y}=\left(x-y\right)+\frac{2}{x-y}\ge2\sqrt{\left(x-y\right)\cdot\frac{2}{x-y}}=2\sqrt{2}.\)

có \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{matrix}\right.\)

=>`x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2>=0`

`<=>2x^2+2y^2+2z^2>=2xy+2yz+2zx`

`<=>x^2+y^2+z^2>=xy+yz+zx`

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:\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\left(dpcm\right)\)
 

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

\(P\ge\frac{2\left(1+x^2\right)}{3+y^2+2z^2}+\frac{2\left(1+y^2\right)}{3+z^2+2x^2}+\frac{2\left(1+z^2\right)}{3+x^2+2y^2}\)

Đặt \(\left\{{}\begin{matrix}3+y^2+2z^2=a\\3+z^2+2x^2=b\\3+x^2+2y^2=c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}1+x^2=\frac{c+4b-2a}{9}\\1+y^2=\frac{a+4c-2b}{9}\\1+z^2=\frac{b+4a-2c}{9}\end{matrix}\right.\) với \(a;b;c\ge3\)

\(\Rightarrow P\ge\frac{2\left(c+4b-2a\right)}{9a}+\frac{2\left(a+4c-2b\right)}{9b}+\frac{2\left(b+4a-2c\right)}{9c}\)

\(\Rightarrow P\ge\frac{2}{9}\left(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}\right)+\frac{8}{9}\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)-\frac{4}{3}\)

\(\Rightarrow P\ge\frac{2}{9}.3+\frac{8}{9}.3-\frac{4}{3}=2\)

Dấu "=" xảy ra khi \(a=b=c\) hay \(x=y=z=1\)