# Sum of GCD and LCM

If $a,b \in \mathbb{N}$ and $ab > 2$ show that:

$$\text{lcm}(a, b) + \gcd(a, b) \le ab + 1$$

Let the lcm be $l$ and let the gcd be $g$.

We have to show:

$$g + l \le ab + 1$$

I know that:

$$gl = ab > 2$$

WLOG, suppose $a > b$

I have seen that: $g \le b$ and $l \ge b$ But that doesnt help

• The constraint $ab>2$ is completely unnecesary. This holds for $(a,b)=(1,1)$ and for $\{a,b\}=\{1,2\}$ too. I am assuming $0\not\in\mathbb N$. – user26486 Mar 22 '15 at 17:46
• Even if $0\in\mathbb N$ in your context, we would still have a stronger statement if $ab\ge 1$ instead of $ab>2$. – user26486 Mar 22 '15 at 19:12

Since $gl = ab$, we have $$ab + 1 = gl + 1 = g + l + gl - g - l + 1 = g + l +(g - 1)(l - 1) \ge g + l.$$
• The $ab>2$ constraint is completely unnecessary. The inequality holds without the constraint (assuming OP meant $0\not\in\mathbb N$) and your answer doesn't use it (again, assuming $0\not\in\mathbb N$). – user26486 Mar 22 '15 at 17:48
• Though it is necessary if we assume $0\in\mathbb N$. Peano axioms wikipedia page includes it and Natural number wikipedia page includes it as the common convention. – user26486 Mar 22 '15 at 18:03
• @Joffan $\gcd(a,b),a,b\in\mathbb N_{0}$ is not well defined iff $a=b=0$. $\gcd(a,0)=a,\forall a\in\mathbb N_{\ge 1}$. So it is defined if you allow at most one zero, just not two (when talking about $\gcd$ of two numbers). – user26486 Mar 22 '15 at 19:10
• Instead of assumptions, it would probably be better if the OP would be edited to use $\mathbb{N}^+$ or $\mathbb{N}^0$. – nyuszika7h Mar 22 '15 at 22:11