Have: n³ - 3n² + 2 = n³ - n² - 2n² + 2

= n²(n - 1) - 2n² + 2

It is easy to see that n²(n - 1) is an even number because it contains n(n - 1) which is the product of two consecutive natural numbers, so n²(n - 1) - 2n² + 2 is an even number

But n³ - 3n² + 2 = n²(n - 1) - 2n² + 2 is a prime number so n³ - 3n² + 2 = 2

=> n³ - 3n² = 0 

<=> n²(n - 3) = 0

According to the topic, n should be positive n = 3

Set x + 29 = t       (*)

The given equation becomes: \(\dfrac{1}{t^2}+\dfrac{1}{\left(t+1\right)^2}=\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{\left(t+1\right)^2+t^2}{t^2\left(t+1\right)^2}=\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{t^2+2t+1+t^2}{t^2\left(t^2+1+2t\right)}=\dfrac{5}{4}\)

\(\Leftrightarrow4\left(t^2+2t+1+t^2\right)=5t^2\left(t^2+1+2t\right)\)

\(\Leftrightarrow\) 8t² + 8t + 4 = 5t⁴ + 5t² + 10t³

\(\Leftrightarrow\) 5t⁴ + 10t³ - 3t² - 8t - 4 = 0

\(\Leftrightarrow\) 5t⁴ - 5t³ + 15t³ - 15t² + 12t² - 12t + 4t - 4 = 0

\(\Leftrightarrow\left(t-1\right)\left(5t^3+15t^2+12t+4\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(5t^3+10t^2+5t^2+10t+2t+4\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+2\right)\left(5t^2+5t+2\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+2\right).5.\left[\left(t+\dfrac{5}{4}\right)^2-\dfrac{93}{80}\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}t-1=0\\t+2=0\\\left(t+\dfrac{5}{4}\right)^2-\dfrac{93}{80}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-2\\\left(t+\dfrac{5}{4}\right)^2=\dfrac{93}{80}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-2\\t=\dfrac{-25+\sqrt{465}}{20}\\t=\dfrac{-25-\sqrt{465}}{20}\end{matrix}\right.\)

Now just replace (*) is ok

+ With n = 0, instead we have: 0⁴ + 8.0³ + 19.0² - 33.0 - 90 = 0

<=> -90 = 0, nonsense

+ With n = 1, instead we have: 1⁴ + 8.1³ + 19.1² - 33.1 - 90 = 0

<=> 1 + 8 + 19 - 33 - 90 = 0

<=> -95 = 0, nonsense

+ With n = 2, instead we have: 2⁴ + 8.2³ + 19.2² - 33.2 - 90 = 0

<=> 16 + 64 + 76 - 66 - 90 = 0

<=> 0 = 0, always true

+ With \(n>2\), we have:

\(n^4+8n^3+\dfrac{5}{2}n^2>2^4+8.2^3+\dfrac{5}{2}.2^2=90\)

\(\dfrac{33}{2}n^2>\dfrac{33}{2}.2n=33n\)

The two sides of the two inequalities above are:

\(n^4+8n^3+19n^2>90+33n\)

\(\Leftrightarrow n^4+8n^3+19n^2-33n-90>0\), contrary to the assumption

So n = 2

\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x}.\dfrac{y+1}{y}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x}=\dfrac{3}{2}:\dfrac{y+1}{y}=\dfrac{3y}{2y+2}\)

\(\Leftrightarrow\left(x+1\right)\left(2y+2\right)=3xy\)

\(\Leftrightarrow2xy+2y+2x+2=3xy\)

\(\Leftrightarrow2y+2x+2-xy=0\)

\(\Leftrightarrow2\left(x-2\right)-y\left(x-2\right)+6=0\)

\(\Leftrightarrow\left(2-x\right)\left(y-2\right)=-6\)

\(\Leftrightarrow\left(x-2\right)\left(y-2\right)=6\)

\(\Rightarrow6⋮x-2\) 

But x;y be positive integer should \(x-2\ge-1;y-2\ge-1\)

\(\Rightarrow x-2\in\left\{1;2;3;6\right\}\)

\(\Rightarrow x\in\left\{3;4;5;8\right\}\)

The corresponding value of y is: 8; 5; 4; 3

a) First of all, we prove sub-equality: \(\dfrac{1}{\left(1+x\right)^2}+\dfrac{1}{\left(1+y\right)^2}\ge\dfrac{1}{1+xy}\)

\(\Leftrightarrow\left(1+xy\right)\left[\left(1+y\right)^2+\left(1+x\right)^2\right]\ge\left(1+x\right)^2\left(1+y\right)^2\)

\(\Leftrightarrow\left(1+xy\right)\left(1+y^2+2y+1+x^2+2x\right)\ge\left[\left(1+x\right)\left(1+y\right)\right]^2\)

\(\Leftrightarrow\left(1+xy\right)\left[2\left(1+x+y\right)+x^2+y^2\right]\ge\left[1+x+y+xy\right]^2\)

\(\Leftrightarrow2\left(1+xy\right)\left(1+x+y\right)+\left(1+xy\right)\left(x^2+y^2\right)\)\(\ge\left(1+x+y\right)^2+x^2y^2+2\left(1+x+y\right)xy\)

\(\Leftrightarrow2xy\left(1+x+y\right)+2\left(1+x+y\right)+x^2+y^2+xy\left(x^2+y^2\right)\ge1+x^2+y^2+2\left(x+xy+y\right)+x^2y^2+2\left(1+x+y\right)xy\)

\(\Leftrightarrow1+xy\left(x^2+y^2\right)\ge2xy+x^2y^2\)

\(\Leftrightarrow xy\left(x^2+y^2-2xy\right)+\left(x^2y^2-2xy+1\right)\ge0\)

\(\Leftrightarrow xy\left(x-y\right)^2+\left(xy-1\right)^2\ge0\), always true for x;y positive

Apply to my article:

\(\left(\dfrac{1}{1+a}\right)^2+\left(\dfrac{1}{1+b}\right)^2+\left(\dfrac{1}{1+c}\right)^2\ge\)\(\dfrac{1}{1+ab}+\left(\dfrac{1}{1+c}\right)^2\)

                                                                   \(\ge\dfrac{1}{1+\dfrac{1}{c}}+\dfrac{1}{\left(1+c\right)^2}=\dfrac{c}{c+1}+\dfrac{1}{\left(c+1\right)^2}\)

                                                                   \(\ge\dfrac{c^2+c+1}{c^2+2c+1}\)

Need to prove: \(\dfrac{c^2+c+1}{c^2+2c+1}\ge\dfrac{3}{4}\)

\(\Leftrightarrow4c^2+4c+4\ge3c^2+6c+3\)

\(\Leftrightarrow\)\(c^2-2c+1\ge0\)

\(\Leftrightarrow\left(c-1\right)^2\ge0\)

The "=" sign occurs when a = b = c = 1