a+b+c = de
The sum of three 1-digit integers is a 2-digit integer. The tens digit of that 2-digit integer MIGHT be 2; but it is much more likely to be 1. So let's assume d=1.
de+d+e = fg
If d=1, then this becomes 1e+1+e = fg, or 11+2*e = fg. That means f has to be 2. So now we know 11+2*e = 2g.
fg+f+g = hi
Since f=2, this becomes 2g+2+g = hi, or 22+2*g = hi. That means h has to be 3. So now we know 22+2*g = 3i.
So digits 1, 2, and 3 are d, f, and h; so a, b, c, e, g, and i must be, in some order, digits 4 through 9.
Now we now 11+2*e = 2g; and g can't be 1, 2, or 3. That means e has to be 7, 8, or 9.
And we know 22+2*g = 3i; and i can't be 1, 2, or 3. That means g has to be 6, 7, 8, or 9.
That's all the logical analysis I can see that we can do. Now we just need to try sets of digits for a, b, and c for which the sum a+b+c is greater than 13 and less than 20; and see where that leads us. For each set of digits we choose for a, b, and c, we stop looking at that case when a letter needs to be a digit that has already been used.
a, b, c e (units digit g (units digit i (units digit
of a+b+c) of 11+2*e) of 22+2*g)
-----------------------------------------------------------
456 5 X
457 6 3 X
458 7 5 X
459 8 7 6 << YES; this should be the solution; but let's keep checking
467 7 X
468 8 X
469 9 X
478 9 9 X
567 8 7 X
568 9 9 X
There is indeed a single solution: (a,b,c,d,e,f,g,h,i) = (9,5,4,1,8,2,7,3,6)
a>b>c? 9>5>4? YES
a+b+c = de? 9+5+4 = 18? YES
de+d+e=fg? 18+1+8 = 27? YES
fg+f+g=hi? 27+2+7 = 36? YES