Ta có:

\(\frac{2.\left(x^2+x+1\right)}{x^2+1}=\frac{2.\left(x^2+1\right)+2x}{x^2+1}=2+\frac{2x}{x^2+1}\)

Ta có:\(2+\frac{2x}{x^2+1}-1=1+\frac{2x}{x^2+1}\)

\(=\frac{x^2+2x+1}{x^2+1}=\frac{\left(x+1\right)^2}{x^2+1}\ge0\)  \(\Rightarrow\frac{2.\left(x^2+x+1\right)}{x^2+1}\ge1\)

\(2+\frac{2x}{x^2+1}-3=\frac{2x}{x^2+1}-1=\frac{-x^2+2x-1}{x^2+1}\)

\(=\frac{-\left(x-1\right)^2}{x^2+1}\le0\) \(\Rightarrow\frac{2.\left(x^2+x+1\right)}{x^2+1}\le3\)

Vậy \(1\le\frac{2.\left(x^2+x+1\right)}{x^2+1}\le3\)

\(\frac{1}{3}< =\frac{x^2+x+1}{x^2-x+1}\Rightarrow x^2-x+1< =3x^2+3x+3\Rightarrow x^2-x+1-3x^2-3x-3< =0\)

\(\Rightarrow-2x^2-4x-2< =0\Rightarrow-2\left(x^2+2x+1\right)< =0\Rightarrow-2\left(x+1\right)^2< =0\)

vì \(\left(x+1\right)^2>=0;-2< 0\Rightarrow-2\left(x+1\right)^2< =0\)luôn đúng \(\Rightarrow\frac{1}{3}< =\frac{x^2+x+1}{x^2-x+1}\)luôn dúng (1)

cái kia cx tương tự như vậy nhé

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

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

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

\(\le3+\frac{z^2}{2xy}+\frac{x^2}{2yz}+\frac{y^2}{2zx}\)\(=3+\frac{x^3+y^3+z^3}{2xyz}\)

Dấu "=" \(\Leftrightarrow x=y=z=\frac{\sqrt{3}}{3}\)

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=\(\frac{1}{2}\)

5/ Tưỡng dễ ăn = sos + bđt phụ ai ngờ....hic...

\(BĐT\Leftrightarrow\Sigma_{cyc}\left(\frac{a^2+b^2+c^2}{a+b+c}-\frac{a^2+b^2}{a+b}\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\left(\frac{\left(a^2+b^2+c^2\right)\left(a+b\right)-\left(a^2+b^2\right)\left(a+b+c\right)}{\left(a+b+c\right)\left(a+b\right)}\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\frac{ca\left(c-a\right)-bc\left(b-c\right)}{\left(a+b+c\right)\left(a+b\right)}\ge0\)\(\Leftrightarrow\Sigma_{cyc}\left(\frac{ca\left(c-a\right)}{\left(a+b+c\right)\left(a+b\right)}-\frac{ca\left(c-a\right)}{\left(a+b+c\right)\left(b+c\right)}\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\frac{ca\left(c-a\right)^2}{\left(a+b+c\right)}\ge0\left(\text{đúng}\right)\)

Ai ngờ nổi khi không dùng BĐT phụ lại dễ hơn cái kia chứ -_-

Ây za,nhầm dòng cuối cùng xíu ạ:

\(\Leftrightarrow\Sigma_{cyc}\frac{ca\left(c-a\right)^2}{\left(a+b+c\right)\left(a+b\right)\left(b+c\right)}\ge0\left(\text{đúng}\right)\) -_- đánh thiếu một chút lại ra nông nỗi -_-

Bài 1:

Xét các hiệu sau:

\(M=x^3+y^3+z^3-(x^2+y^2+z^2)=x^2(x-1)+y^2(y-1)+z^2(z-1)\)

\(N=x^4+y^4+z^4-(x^3+y^3+z^3)=x^3(x-1)+y^3(y-1)+z^3(z-1)\)

Lấy $N-M$:

\( N-M=\sum x^2(x-1)(x-1)=\sum x^2(x-1)^2\geq 0\)

\(\Leftrightarrow \sum x^4-2\sum x^3+\sum x^2\geq 0\)

\(\Rightarrow \sum x^4\geq 2\sum x^3-\sum x^2(*)\)

\(P=x^5+y^5+z^5-(x^4+y^4+z^4)=x^4(x-1)+y^4(y-1)+z^4(z-1)\)

Lấy $P-M$

\(P-M=\sum x^2(x-1)(x^2-1)=\sum x^2(x-1)^2(x+1)\geq 0, \forall x,y,z>-1\)

\(\Leftrightarrow \sum x^5-\sum x^4-\sum x^3+\sum x^2\geq 0\)

\(\Leftrightarrow \sum x^5\geq \sum x^4+\sum x^3-\sum x^2\). Kết hợp với (*) và điều kiện ban đầu suy ra:

\(\sum x^5\geq 2\sum x^3-\sum x^2+\sum x^3-\sum x^2=3\sum x^3-2\sum x^2\geq \sum x^2\)

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

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) và (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 <=> \(x=y=\frac{1}{2}\)