Define f(x) as the LHS of the equation. Clearly f(x) is continuous.

For sufficiently large  or small x, x4 will always dominate the other terms, and is of course always positive, so as x approaches +- infinity f(x) is positive.

Now, suppose b > -¼. Then -b -¼ is less than zero and hence f(0) < 0. If f -> infinity as x -> + or - infinity and f(0) < 0 then the function must cross the x-axis between 0 and +infinity, and also between 0 and -infinity. So there must be at least one real root each where x < 0 and where x > 0. 

Indeed, if we can find any f(k) < 0 then the curve must cross the x-axis at least twice and so f(x) must have at least two real roots. So let’s determine the conditions under which f(k) > 0 for all k, as this is the only way for f(x) to have no real roots.

Consider f(n) and f(-n). If these are both > 0 then so is their sum.

f(n) = n^4 + an^3  +2bn^2 + cn -b -¼ and 

f(-n) = n^4 - an^3  +2bn^2 - cn -b -¼ and their sum is

2n^4 + 4bn^2 - 2b - ½

= 4n^4 + 8bn^2 -4b -1

= 4b(2n^2 - 1) + (2n^2 - 1)(2n^2 + 1) > 0

Then b > - (2n^2 - 1)(2n^2 + 1) / 4(2n^2 -1) = -(2n^2 + 1)/4

Since this is true for any n, it’s true in the limit that n -> 0 where we get b > -1/4

Well b can’t be both > -¼ AND < -¼ so the assumption that f(x) > 0 for all x must be false, and there must be at least one k for which f(k) < 0 AND as a result there must be at least two real roots for f(x).

One might argue that there’s a potential loophole if b = -¼ exactly, but in this case x = 0 is a real root for f(x) and so even here f(x) has a real root.

But what if the problem meant that we should prove the equation has ONLY real roots?

Consider the special case where a = b = c = 0 which results in f(x) = x4 - ¼

f(x) = 0 holds when x = +- sqrt(2) and also when x = +- sqrt(2)i This case clearly does have some real roots, but equally clearly doesn’t have ALL real roots, so it is NOT the case that f(x) has all real roots for all values of (a,b,c)