chứng minh \(\frac{3}{2}\ge\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\)

ta có \(\left(x-1\right)^2\ge0\Leftrightarrow x^2+1\ge2x\Leftrightarrow\frac{2x}{1+x^2}\le1\)

\(\left(y-1\right)^2\ge0\Leftrightarrow y^2+1\ge2y\Leftrightarrow\frac{2y}{1+y^2}\le1\)

\(\left(z-1\right)^2\ge0\Leftrightarrow z^2+1\ge2z\Leftrightarrow\frac{2z}{1+z^2}\le1\)

\(\Rightarrow\frac{2x}{1+x^2}+\frac{2y}{1+y^2}+\frac{2x}{1+z^2}\le3\Leftrightarrow\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\le\frac{3}{2}\)

chứng minh \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge\frac{3}{2}\)

áp dụng bất đẳng thức Cauchy ta có: 

\(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge3\sqrt[3]{\frac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}=\frac{3}{\sqrt{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}\)

ta lại có \(\frac{\left(1+x\right)\left(1+y\right)\left(1+z\right)}{3}\ge\sqrt[3]{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)

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

kết hợp ta có \(\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\le\frac{3}{2}\le\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\)

Đặt \(x=a+1;y=b+1;z=c+1\Rightarrow0\le a,b,c\le2\)và \(a+b+c=3\)

Chứng minh : \(\left(a+1\right)^3+\left(b+1\right)^3+\left(c+1\right)^3\le36\)

\(\Leftrightarrow a^3+b^3+c^3+3\left(a^2+b^2+c^2\right)\le24\). Không mất tính tổng quát, giả sử \(2\ge a\ge b\ge c\ge0\) thì:

\(3a\ge a+b+c=3\Rightarrow2\ge a\ge1\Rightarrow\left(a-1\right)\left(a-2\right)\le0\)

Theo kết quả bài này thì \(a^2+b^2+c^2\le5\) (em làm thế này cho ngắn, lúc trình bày vô bài làm thì anh ghi cả chứng minh vô luôn nha!). Vậy ta chỉ cần chứng minh: \(a^3+b^3+c^3\le9\).

Ta có: \(a^3+b^3+c^3\le a^3+b^3+c^3+3bc\left(b+c\right)\)

\(=a^3+\left(b+c\right)^3=a^3+\left(3-a\right)^3\)

\(=9\left(a-1\right)\left(a-2\right)+9\le9\)

Đẳng thức xảy ra khi \(\left(a;b;c\right)=\left(2;1;0\right)\) và các hoán vị.

Ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)=4\)

=> \(\orbr{\begin{cases}x+y+z=2\\x+y+z=-2\end{cases}}\)

+ \(x+y+z=2\)

Thay vào Pt (1)

=> \(xy+z\left(2-z\right)=1\)

 => \(xy=\left(z-1\right)^2\)=> \(x,y,z\ge0\)( do \(x+y+z=2>0\))

Mà \(xy\le\left(\frac{x+y}{2}\right)^2=\left(\frac{2-z}{2}\right)^2\)

=> \(z-1\le\frac{2-z}{2}\)=> \(z\le\frac{4}{3}\)

Hoàn toàn TT => \(x,y,z\le\frac{4}{3}\)

+ \(x+y+z=-2\)

=> \(xy+z\left(-2-z\right)=1\)

=> \(xy=\left(z+1\right)^2\)=> \(x,y,z\le0\)( do \(x+y+z=-2< 0\))

Mà \(xy\le\left(\frac{x+y}{2}\right)^2=\left(\frac{-2-z}{2}\right)^2\)

=> \(\left(z+1\right)^2\le\left(\frac{z+2}{2}\right)^2\)

=> \(z+1\ge\frac{-z-2}{2}\)=> \(z\ge-\frac{4}{3}\)

TT => \(x,y,z\ge-\frac{4}{3}\)

Vậy \(-\frac{4}{3}\le x,y,z\le\frac{4}{3}\)

Bài 1: Áp dụng BĐT AM-GM ta có:

\(1+x\ge2\sqrt{x}\)

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

\(y+1\ge2\sqrt{y}\)

Cộng theo vế 3 BĐT trên ta có:

\(2\left(1+x+y\right)\ge2\left(\sqrt{x}+\sqrt{xy}+\sqrt{y}\right)\)

\(1+x+y\ge\sqrt{x}+\sqrt{xy}+\sqrt{y}\Leftrightarrow VT\ge VP\) 

Đẳng thức xảy ra khi  \(\hept{\begin{cases}1+x=2\sqrt{x}\\x+y=2\sqrt{xy}\\y+1=2\sqrt{y}\end{cases}}\Rightarrow x=y=1\)

Khi đó \(S=x^{2013}+y^{2013}=1^{2013}+1^{2013}=2\)

Bài 2: Vì \(\hept{\begin{cases}x,y,z\in\left[-1;3\right]\\x+y+z=3\end{cases}}\) nên 

\(0\le\left(x+1\right)\left(y+1\right)\left(z+1\right)+\left(3-x\right)\left(3-y\right)\left(3-z\right)\)

\(\Leftrightarrow0\le4\left(xy+yz+xz\right)-8\left(x+y+z\right)+28\)

\(\Leftrightarrow0\le2\left(xy+yz+xz\right)+2\)

\(\Leftrightarrow x^2+y^2+z^2\le x^2+y^2+z^2+2\left(xy+yz+xz\right)+2\)

\(\Leftrightarrow x^2+y^2+z^2\le\left(x+y+z\right)^2+2\)

\(\Leftrightarrow x^2+y^2+z^2\le3^2+2=9+2=11\)

Cảm ơn b Thắng Nguyễn