\( x^3=a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}+a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}+3\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}.\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}.x\)

=> \(x^3=2a+3\sqrt[3]{\left(a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)\left(a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)}.x\)

\(x^3=2a+3\sqrt[3]{a^2-\left(\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)^2}.x\)

\(x^3=2a+3\sqrt[3]{\left(\frac{1-2a}{3}\right)^3}.x\)=> \(x^3=2a+\left(1-2a\right).x\)

=> x³   = 2a + x - 2ax => x³ - x + 2ax - 2a = 0 

=> x(x²  - 1) + 2a.(x -1) = 0 

=> (x -1). (x² + x + 2a) = 0 

=> x - 1 = 0 hoặc x² + x  + 2a = 0 

Mà x² + x + 2a = x² + 2.x . (1/2) + (1/4) + 2a -(1/4) = (x +1/2)² + 2. (a - 1/8) > = 0 với mọi a > = 1/8

=>  x² + x  + 2a = 0  Vô nghiệm

vậy x = 1 => x thuộc N

Dấu ở giữa là cộng chứ nhỉ??

Đặt \(y=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}};z=\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\)

\(\Rightarrow\hept{\begin{cases}y^3+z^3=2a\\yz=\sqrt[3]{a^2-\frac{\left(a+1\right)^2\left(8a-1\right)}{27}}\\y+z=x\end{cases}=\sqrt[3]{\frac{27a^2-\left(8a^3+15a^2+6a-1\right)}{27}}=\sqrt[3]{\frac{\left(1-2a\right)^3}{27}}=\frac{1-2a}{3}}\)

Thay vào ta được:

\(x^3=\left(y+z\right)^3=y^3+z^3+3yz\left(y+z\right)\)\(=2a+3\frac{1-2a}{3}x=2a+\left(1-2a\right)x\)

\(\Leftrightarrow x^3-\left(1-2a\right)x-2a=0\)

\(\Leftrightarrow x^3-x+2ax-2a=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x^2+2a+x=0\end{cases}}\)

Đến đây thì có lẽ là sẽ cm được \(x^2+2a+x>0\), mình chưa tìm ra cách cm.

KL : \(x=1\inℤ\)

\(x^3=2a+3x.\sqrt[3]{a^2-\frac{\left(a+1\right)^2}{9}\left(\frac{8a-1}{3}\right)}\)

\(x^3=2a+3x\sqrt[3]{\frac{1-6a+12a^2-8a^3}{27}}\)

\(x^3=2a+3x\sqrt[3]{\left(\frac{1-2a}{3}\right)^3}\)

\(x^3=2a+\left(1-2a\right)x\)

\(x^3-x+2ax-2a=0\)

\(x\left(x^2-1\right)+2a\left(x-1\right)=0\)

\(\left(x-1\right)\left(x^2+x+2a\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x^2+x+2a=0\left(1\right)\end{matrix}\right.\)

Xét (1): \(x^2+x+\frac{1}{4}+2a-\frac{1}{4}=0\Rightarrow\left(x+\frac{1}{2}\right)^2+2\left(a-\frac{1}{8}\right)=0\)

Do \(a>\frac{1}{8}\Rightarrow\left(x+\frac{1}{2}\right)^2+2\left(a-\frac{1}{8}\right)>0\)

\(\Rightarrow\left(1\right)\) vô nghiệm \(\Rightarrow x=1\) hay x nguyên dương

ap dung bdt am gm

\(\sqrt{1+8a^3}=\sqrt{\left(1+2a\right)\left(4a^2-4a+1\right)}\)\(\le\frac{1+2a+4a^2-2a+1}{2}=\frac{4a^2+2}{2}=2a^2+1\)

\(\Rightarrow\frac{1}{\sqrt{1+8a^3}}\ge\frac{1}{2a^2+1}\)

tuongtu ta cung co \(\frac{1}{\sqrt{1+8b^3}}\ge\frac{1}{2b^2+1};\frac{1}{\sqrt{1+8c^3}}\ge\frac{1}{2c^2+1}\)

\(\Rightarrow\)VT\(\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\)

tiep tuc ap dung bat cauchy-schwarz dang engel ta co

\(VT\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\ge\frac{\left(1+1+1\right)^2}{2\left(a^2+b^2+c^2\right)+3}=\frac{3^2}{6+3}=1\)(dpcm)

dau = xay ra \(\Leftrightarrow a=b=c=1\)