dùng đt

\(P=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)

\(\Leftrightarrow P^3=1+\frac{\sqrt{84}}{9}+1-\frac{\sqrt{84}}{9}+\sqrt[3]{\left(1+\frac{\sqrt{84}}{9}\right)\left(1-\frac{\sqrt{84}}{9}\right)}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)\)

\(\Leftrightarrow P^3=2-P\)

\(\Leftrightarrow\left(P-1\right)\left(P^2+P+2\right)=0\)

\(\Leftrightarrow P=1\).

Sửa đề: \(B=\frac{x}{1+y^2}+\frac{y}{1+z^2}+\frac{z}{1+x^2}\)

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

\(\ge x-\frac{xy^2}{2y}=x-\frac{xy}{2}\left(Cauchy\right)\)

Tương tự CM được: \(\frac{y}{1+z^2}\ge y-\frac{yz}{2}\) ; \(\frac{z}{1+z^2}\ge z-\frac{zx}{2}\)

Cộng vế 3 BĐT trên lại với nhau ta được:

\(B\ge\left(x+y+z\right)-\frac{xy+yz+zx}{2}\)

\(\ge3-\frac{\frac{\left(x+y+z\right)^2}{3}}{2}=3-\frac{3}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi: \(x=y=z=1\)

R= Đường ngang vạch như đoạn elip vậy Ta xét R=\(\eta\)

\(x^{12}-x^9+x^4-x+1>0\)\(\Leftrightarrow2x^{12}-2x^9+2x^4-2x+2>0\)

\(\Leftrightarrow\left(x^{12}-2x^9+x^6\right)+\left(x^{12}-x^6+\frac{1}{4}\right)+\left(2x^4-2x^2+\frac{1}{2}\right)+\)\(\left(2x^2-2x+\frac{1}{2}\right)+\frac{3}{4}>0\)

\(\Leftrightarrow\left(x^6-x^3\right)^2+\left(x^6-\frac{1}{2}\right)^2+2\left(x^2-\frac{1}{2}\right)^2+2\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

do đó ta có đpcm

\(D=x^{10}-x^9+x^4-x+1>0\)

\(D=x^9\left(x-1\right)+x\left(x^3-1\right)+1\)

Vậy ta xét : \(x\ge1\)\(\Rightarrow\)D Sẽ luôn dương (1)

Xét: \(x< 1\)

\(\Rightarrow\)\(D=x^{10}+x^4\left(1-x^5\right)+\left(1-x\right)\)

ko làm đc