It's not that trivial if we really insist on
all the details. Let's assume x > 0 throughout. We can take sqrt(x) < x as a freebie.
An important lemma we need is that if f(x) is continuous on [a,b] and f'(x) > 0 on (a,b), then f(a) < f(b). For this, we need to write out the mean value thoerem, which needs Rolle's theorem, which needs the extreme value theorem, which gets into the definition of the real numbers. As well as all the fun epsilon–delta arguments for the limit manipulations, of course.
x/exp(sqrt(4x+3)) < x thankfully follows trivially from exp(sqrt(4x+3)) > 1 = exp(0). Since sqrt(4x+3) > 0, we just need to show that exp(z) is increasing. For this, we can treat the exponential as the anti-logarithm (since that's how my high school textbook did it), then show that log(z) is increasing, which follows from log'(z) = 1/z > 0 and our lemma.
For log(x^2+1) < x, we'll want to use our lemma on f(z) = z - log(z^2+1) to show that f(x) > f(0) = 0. Here, f'(z) = 1 - 2x/(x^2+1), for which we need the chain rule and the power rule (or a specialized epsilon–delta argument). Since x^2+1 > 0, f'(z) > 0 is implied by (x^2+1) - 2x > 0, which luckily factors into the trivial (x-1)^2 > 0. So ultimately, we break the lemma up into (0,1) and (1,x) to avoid trouble with the stationary point.
The trouble with problems like these is the whole foundation that the obvious lemmas have to be built upon. "Just look at the graph, of course it's increasing" isn't a rigorous proof. Of course, if you want to do this seriously, then you go and build that foundation, and on top of that you probably define some helpful lemmas for the ladder of asymptotic growth rates. But all of the steps must be written out at some point or another.