How to prove x + square root(x²+1) > 0

sqrt(x^2+1) is positive for all real x and is >sqrt(x^2)=|x|
so x+sqrt(x^2+1)>x+|x|
Even if x is negative x+|x|>=0 so result follows.

Note that if you start from the hypothesis and get a true result it does
not follow that the hypothesis is correct. A simple example is the
hypothesis : 1=2
Follows that 2=1
Now add to give 3=3 and this is true but you have not proved that 1=2

Well lets consider the case where there is no +1.
The square root of x^2 is x (we disregard -x as x^2 has to be positive).
If x is +ve, then the answer must be +ve (+ve + +ve)

If the x is -ve, then then answer is 0, as we have a -ve number plus it's opposite (+ve).
So if we put in the +1, the square root will be slightly higher than the -ve x, so we have x+a-x = a,
where a is the small positive bit added on by the +1 (It should be noted that this value is NOT 1)

since √(x^2) = x
hence √[(x^2)+1] > x it is always true.
also √[(x^2)+1] > 0 for all real x even if x is -ve.
so x + √[(x^2)+1] > 0

hypothesis: X + square root(x²+1) > 0
then: square root(x²+1) > X
x²+1>x²
1>0
So this hypothesis is correct : x + square root(x²+1) > 0