Here is an answer giving the outline of a proof.
The Bound is tight
If the bound is true then it is tight: If $a=1$, $b=1$ then
$\mathbb{P}[X\geq \mathbb{E}[x]] = \mathbb{P}[X\geq 1] =1-\mathbb{P}[X=0] = 1-\frac{1}{2}\frac{1}{2}=\frac{3}{4}$
Getting a feel for the problem
As a rule of thumb (though can be made precise) if $a>5$ and $b>5$ then we can approximate the Binomial by normal random variable by the central limit theorem. That is $X\sim N(a,Var(X))$ approximately. The error in this approximation becomes increasing small as $a$ and $b$ increase.
Hence $\mathbb{P}[X\geq \mathbb{E}[x]] \approx \frac{1}{2}<\frac{3}{4}$ for $a>5$ and $b>5$
To make precise, first read this: http://en.wikipedia.org/wiki/Binomial_distribution#Normal_approximation
and then: Box, Hunter and Hunter (1978). Statistics for experimenters. Wiley. p. 130
We are left with three cases: a and b both small, a large and b small, b large and a small.
$b$ large and $a\leq 5$
If $b+a>100$ then a poisson approximation to the Binomial r.v. becomes appealling. That is $X\sim Po(a)$ approximately.
Then $\mathbb{P}(X\geq a)=1-\sum_{i=0}^{a-1}\frac{e^{-a}a^i}{i!}\leq 0.63<\frac{3}{4}$
For explicit error bounds look in A bound on the Poisson-binomial relative error by Teerapabolarn (2007). Theorem 2.1 contains a good error bound.
$b\leq 5$ and $a$ large
Let $Y=a+b-X\sim \text{Bin}(a+b,\frac{b}{a+b})$ now if $b+a>100$ $Y$ is approximately Poisson and
$\mathbb{P}(X\geq a)\approx\mathbb{P}(Y\leq b) <0.45 \leq \frac{3}{4}$
Remaining cases
Note that the cases above are not quite mutually exclusive, but there a finite number of cases left. The cases left are
1) $a\leq 5$ and $b\leq 100$
2) $b\leq 5$ and $a\leq 100$
And this can be checked on a computer. To give the bound.