Let $a=da_0$, $b=db_0$, thus $da_0b_0+d=kd(a_0+b_0) \implies k=\frac{a_0b_0+1}{a_0+b_0}$. Assuming to contrary, say $k > \frac{a+b}4 \implies 4a_0b_0+4 > d(a_0^2+b_0^2+2a_0b_0) > a_0^2+b_0^2+2a_0b_0 \implies 4 > (a_0-b_0)^2$. Thus $|a_0-b_0| < 2$. Considering $a_0 \geq b_0$, we get $a_0$ as one of $b_0,b_0+1$. As $k$ is an integer, $\frac{a_0b_0+1}{a_0+b_0}$ is an integer $\implies a_0+b_0 \mid a_0b_0+1$. Case 1: $a_0=b_0$: $2a_0 \mid a_0^2+1 \implies \mid 2a_0^2+2-2a_0^2 \implies \mid 2$. Thus $a_0=b_0=1$, which is a contradiction as $k>1$. Case 2: $a_0=b_0+1$: $2b_0+1 \mid b_0^2+b_0+1 \implies \mid 2b_0^2+2b_0+2 -b_0(2b_0+1) \implies \mid b_0+2 \implies \mid 2b_0+4-2b_0-1 \implies \mid 3$. Thus $b_0=1 \implies a_0=2$, which is a contradiction, as $k>1$. Hence we're done.

We first try to reduce/simplify the problem statement Let $d=gcd(a,b)$. So, $a=dx,b=dy, gcd(x,y)=1$ Then, we get $k=\frac{xy+1}{x+y}$. We will use this later. $a+b=d(x+y)\geq x+y$, as $d\geq 1$ So, we need to prove $x+y\geq\frac{4xy+4}{x+y}\Leftrightarrow (x-y)^2\geq 4$ WLOG Let $a\geq b\implies x\geq y$ So, we are left to prove $x-y\geq 2$ We proceed by contradiction- assume $x-y\leq 1$, i.e.

It's enough to consider $a$ and $b$ coprime. Then, the problem is to show the implication

If $d=(a,b)$ with $a=da_1$ and $b=db_1$, we then have $a_1b_1+1 = k(a_1+b_1)$, yielding $(a_1-k)(b_1-k)=k^2-1$. Now if $a_1\leqslant k$, so do $b_1$, which clearly yields a contradiction (as $k-a_1\leqslant k-1$ and thus $(k-a_1)(k-b_1)\leqslant (k-1)^2<k^2-1$). Thus $a_1,b_1>k$. In particular there exists a positive integer $\ell\mid k^2-1$ such that $a_1=\ell +k$ and $b_1=k+\frac{k^2-1}{\ell}$. But now $a_1+b_1=2k+\ell+\frac{k^2-1}{\ell}\geqslant 2k+2\sqrt{k^2-1}$ by AM-GM inequality. Since $\ell\mid k^2-1$ and for $k>1$ $2k-1<2\sqrt{k^2-1}<2k$, we must have $a_1+b_1\geqslant 4k$ as claimed.

ubermensch wrote: Let $a=da_0$, $b=db_0$, thus $da_0b_0+d=kd(a_0+b_0) \implies k=\frac{a_0b_0+1}{a_0+b_0}$. Assuming to contrary, say $k > \frac{a+b}4 \implies 4a_0b_0+4 > d(a_0^2+b_0^2+2a_0b_0) > a_0^2+b_0^2+2a_0b_0 \implies 4 > (a_0-b_0)^2$. Thus $|a_0-b_0| < 2$. Considering $a_0 \geq b_0$, we get $a_0$ as one of $b_0,b_0+1$. As $k$ is an integer, $\frac{a_0b_0+1}{a_0+b_0}$ is an integer $\implies a_0+b_0 \mid a_0b_0+1$. Case 1: $a_0=b_0$: $2a_0 \mid a_0^2+1 \implies \mid 2a_0^2+2-2a_0^2 \implies \mid 2$. Thus $a_0=b_0=1$, which is a contradiction as $k>1$. Case 2: $a_0=b_0+1$: $2b_0+1 \mid b_0^2+b_0+1 \implies \mid 2b_0^2+2b_0+2 -b_0(2b_0+1) \implies \mid b_0+2 \implies \mid 2b_0+4-2b_0-1 \implies \mid 3$. Thus $b_0=1 \implies a_0=2$, which is a contradiction, as $k>1$. Hence we're done. Why $a_0=b_0$ or $a_0=b_0+1$? Please Explain

A-Thought-Of-God wrote: ubermensch wrote: Let $a=da_0$, $b=db_0$, thus $da_0b_0+d=kd(a_0+b_0) \implies k=\frac{a_0b_0+1}{a_0+b_0}$. Assuming to contrary, say $k > \frac{a+b}4 \implies 4a_0b_0+4 > d(a_0^2+b_0^2+2a_0b_0) > a_0^2+b_0^2+2a_0b_0 \implies 4 > (a_0-b_0)^2$. Thus $|a_0-b_0| < 2$. Considering $a_0 \geq b_0$, we get $a_0$ as one of $b_0,b_0+1$. As $k$ is an integer, $\frac{a_0b_0+1}{a_0+b_0}$ is an integer $\implies a_0+b_0 \mid a_0b_0+1$. Case 1: $a_0=b_0$: $2a_0 \mid a_0^2+1 \implies \mid 2a_0^2+2-2a_0^2 \implies \mid 2$. Thus $a_0=b_0=1$, which is a contradiction as $k>1$. Case 2: $a_0=b_0+1$: $2b_0+1 \mid b_0^2+b_0+1 \implies \mid 2b_0^2+2b_0+2 -b_0(2b_0+1) \implies \mid b_0+2 \implies \mid 2b_0+4-2b_0-1 \implies \mid 3$. Thus $b_0=1 \implies a_0=2$, which is a contradiction, as $k>1$. Hence we're done. Why $a_0=b_0$ or $a_0=b_0+1$? Please Explain I didn't see $|a_0-b_0| < 2$

Nice one Let $a = xd, b = yd$, the given equation becomes $d + xyd = k(xd+yd) \implies k = \frac{xy+1}{x+y}$. Suppose FTSOC that $4k > a+b$ This means that $\frac{4(xy+1)}{x+y} > d(x+y) \ge x+y$ and so $4xy+4 > (x+y)^2 \implies (x-y)^2 < 4$ and so either $x,y$ are same or they differ by $1$ If $x = y$, we need $\frac{x^2+1}{2x}$ to be an integer, which is possible only when $x=y=1$, but this means that $k=1$, which is impossible since we were given $k>1$ if $y=x+1$, we need $\frac{x^2+x+1}{2x+1}$ to be an integer $\implies 2x+1 | x^2 + x + 1 \implies 2x+1 | 4x^2+4x+4$ but since $2x+1|(2x+1)^2 = 4x^2+4x+1$, it means that $2x+1|3$ and so $x=1, y = 2$. But this again means that $k = 1$, which is impossible. Therefore, the given inequality must hold