I found the following exercise in Chartrand's book at the end of chapter 3, and attempted to prove it, but I am not very satisfied because it does not seem legitimate. I have not officially learned the mathematical proofs so kindly please help me.

Exercise 3.45:

Prove that if $\displaystyle a$ and $\displaystyle b$ are two positive integers, then $\displaystyle a^2(b+1)+b^2(a+1) \geq 4ab$

The following is my weak attempt:

Let $\displaystyle P(x,y): a, b \in \mathbb{Z}^+$, and $\displaystyle Q(x,y): a^2(b+1)+b^2(a+1) \geq 4ab$

The algebraic reduction of $\displaystyle Q(x,y):$

$\displaystyle a^2(b+1)+b^2(a+1) \geq 4ab$

$\displaystyle a^2(b+1)+b^2(a+1)-4ab \geq 0$

$\displaystyle (a-b)^2+a^2b+b^2a-2ab \geq 0$

$\displaystyle (a-b)^2+ab(a+b-2) \geq 0$

We can restate the question as $\displaystyle (a,b \in \mathbb{Z}^+, a >0 $ and $\displaystyle b>0) \Rightarrow (a-b)^2+ab(a+b-2) \geq 0$

Proof:

By trichotomy law, we have three cases, namely $\displaystyle a=b, a>b,$ and $\displaystyle a<b.$

Case 1: Assume that $\displaystyle a=b.$ Since $\displaystyle (a-b)^2 = 0,$ $\displaystyle ab(a+b-2) \geq 0$

Case 2: Assume that $\displaystyle a<b$. Since the square of any integer is a positive integer then $\displaystyle (a-b)^2 > 0$ and $\displaystyle ab(a+b-2) \geq 0$. It follows that $\displaystyle (a-b)^2+ab(a+b-2) \geq 0$

Since Case 3 is similar to case 2, we will not repeat. Q.E.D