Finding the smallest value

Sum S = {√[a2 - 7√2a + 49] }+ {√[a2 - √2ab + b2]} + {√[b2 - 10b + 50]}
dS / da = (1/2)(2a - 7√2) / {√[a2 - 7√2a + 49] } + (1/2)( 2a - √2b) / {√[a2 - √2ab + b2]} … (1) [Note: dS/da is partial derivative only]
dS / db = (1/2)( - √2a + 2b) / {√[a2 - √2ab + b2]} + (1/2)(2b – 10) / {√[b2 - 10b + 50]} … (2)
Equate dS / da = 0 for extreme values =>
(1/2)(2a - 7√2) / {√[a2 - 7√2a + 49] } + (1/2)( 2a - √2b) / {√[a2 - (√2)ab + b2]} = 0
(1/2)(2a - 7√2) / {√[a2 - 7√2a + 49] } = -(1/2)( 2a - √2b) / {√[a2 - √2ab + b2]}
Squaring both sides
(4a2 - 28√2a + 98) / (a2 - 7√2a + 49) = (4a2 - 4√2ab + 2b2) / (a2 - (√2)ab + b2)
[4(a2 - 7√2a + 49) – 98] (a2 - (√2)ab + b2) = [4(a2 - √2ab +b2) – 2b2] (a2 - 7√2a + 49)
98(a2 - (√2)ab + b2) = 2b2(a2 - (7√2)a + 49)
49a2 - 49√2ab + 49b2 = a2b2 - 7√2ab2 + 49b2
49a2 – b2a2 = 49√2ab - 7√2ab2
a2(7 – b)(7+b) = 7√2ab(7 – b)
a(7 – b)[a(7+b) - 7√2b] = 0
a = 0 or b = 7 or a = 7√2b / (7+b) … (3)
Both a = 0 or b = 7 have to be rejected since putting them back to (1) does not give zero.
Equate dS / db = 0 for extreme values =>
(1/2)( - √2a + 2b) / {√[a2 - √2ab + b2]} + (1/2)(2b – 10) / {√[b2 - 10b + 50]} = 0
Following similar simplification yields b = 10a / (5√2 + a)
Putting this into (3) gives a = [(7√2)(10a) / (5√2 + a)] / [7 + 10a / (5√2 + a)]
a = 70√2a / (35√2 + 7a + 10a)
35√2a + 17a2 = 70√2a
a = 0 (rejected) or a = 35√2/17 = 2.9116
Sub into (3) gives b = 35/12 = 2.91667
These values give a sum S exactly equal to13.
By varying a and b a little more positive or negative will show that this is the smallest value.