$a,b\in\mathbb{N}$ and $ab>2$.Suppose,lcm$(a,b)=L,\gcd(a,b)=G$ and $a+b\mid L+G$.Prove that $\dfrac{(a+b)}{4}(a+b)\ge (L+G)$ and $a,b$ are two odd... $a,b\in\mathbb{N}$ and $ab>2$.Suppose,$\text{lcm}(a,b)=L,\gcd(a,b)=G$ and $a+b\mid L+G$.Prove that $\dfrac{(a+b)}{4}\cdot(a+b)\ge (L+G)$. Also prove that equality occurs when $a,b$ are consecutive odd integers.  
Hint: There is an exception when $a,b$ are both equal to 2.
I could not approach this problem at all. Please help.

Without loss of generality, $a\leq b$.
If $\gcd(a,b)\geq2$, then $G+L\leq a+\frac{ab}2\leq a\frac b2+\frac{ab}2=\frac{4ab}4\leq\frac{(a+b)^2}4$
The inequality is sharp except for $a=b=2$ in which case $G+L=\frac{(a+b)^2}4$.
If $\gcd(a,b)=1$, we have $a+b\mid L+G=ab+1=(a+b)b-b^2+1$, so $${a+b\mid b^2-1=(b+1)(b-1)}\tag{1}$$
We want to prove $L+G=ab+1\leq\frac{(a+b)^2}4\Longleftrightarrow 4ab+4\leq(b+a)^2\Longleftrightarrow4\leq(b-a)^2$.
Surely $a<b$. If $b=a+1$, then by $(1)$ we have $2a+1\mid(a+2)a$, but $\gcd(2a+1,a)=1$, so $2a+1\mid a+2$ which is impossible ($a=1\Longrightarrow b=2$ and $ab=2$).
Therefore $b\geq a+2$, so the inequality holds and is sharp except for $b=a+2$, in that case $b$ must be odd, because $\gcd(a,b)=1$.
