1. ## Proof

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

The following is my weak attempt:

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

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

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

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

Proof:

By trichotomy law, we have three cases, namely
$a=b, a>b,$ and $a

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

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

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

2. The proof I wrote above is logically incorrect. I have got the right way to prove it now. Administrator may close the thread.