Using Chebyshev's inequality to obtain lower bounds Let $X_1$ and $X_2$ be i.i.d. continuous random variables with pdf $f(x) = 6x(1-x), 0<x<1$ and $0$, otherwise.
Using Chebyshev's inequality, find the lower bound of $P\left(|X_1 + X_2-1| \le\frac{1}{2}\right)$
What I did:
Using as Chebyshev's inequality,
$P(|X-\mu|\ge a)\le \frac{\sigma^2}{a^2}$
Where $a=\frac{1}{2}$
Finding the variance: $E[X^2] - (E[X])^2$
$  E[X] = \int_{0}^{1} x6x(1-x) dx$ $=6\left[\frac{x^  3}{3}-\frac{x^4}{4}\right]_0^1=6\left[\frac{1}{3}-\frac{1}{4}\right]=  \frac{6}{12}=\frac{1}{2}$
$  E[X^2]= \int_{0}^{1}x^{2}6x(1-x)dx=6\left[\frac{x^  4}{4}-\frac{x^5}{5}\right]_0^1=6\left[\frac{1}{4}-\frac{1}{5}\right]=\frac{6}{20}$
Therefore, $\sigma^2=\frac{6}{20}-\left(\frac{1}{2}\right)^2=\frac{1}{20}$
Putting in Chebyshev's inequality, 
$\frac{\sigma^2}{a^2} $= $\left[\frac{\frac{1}{20}}{\left(\frac{1}{2}\right)^2}\right]$=$\frac{4}{20}=\frac{1}{5} $
But what we need is $\le \frac{1}{2}$ which we get by $1-\frac{1}{5}=\frac{4}{5}$, 
But the answer is $\frac{3}{5}$
Where am I going wrong?

You dropped a 6. It should be $E[X^2]=6/20=3/10$.
$$P(|X_1+X_2-1|\leq 1/2)=1-P(|X_1+X_2-1|>1/2).$$
$$P(|X_1+X_2-1|>1/2)=P(X_1^2+X_2^2+2X_1X_2-2X_1-2X_2+1>1/4).$$
$$P(|X_1+X_2-1|^2>1/4)\leq \frac{6/10+2(1/2)^2-1-1+1}{1/4}=4/10.$$
Thus 
$$P(|X_1+X_2-1|\leq 1/2)\geq 1-4/10=3/5$$
