I found this as a part of a proof of a lemma of the one sided Chebyshev Inequality in the book "A first course in Probability" by Tapas K. Chandra and Dipak Chatterjee page 387 , for more information about the context, the lemma says:
If $E(X)=\mu$ and $Var(X)=\sigma ^2$ then for $a>0$, $$ P(X \geq \mu +a) \leq f(a) , P(X \leq \mu - a ) \leq f(a) $$ where $f(a) = \frac{\sigma^2}{\sigma^2 + a^2}$.
In the second proof it says, it suffices to prove this for $\mu = 0 $. It assumes $c \geq 0$ after some steps it uses $$\inf \{ E(X+c)^2 / (a+c)^2 : c \geq 0\}$$
And my problem is that it says it is $\mathbf{ straightforward}$ to show that this infimum is attined at $\sigma^2 /a$
First let's calculate expectation, $$E\{\frac{(X+c)^2}{(a+c)^2}\}=\frac{1}{(a+c)^2}E\{(X+c)^2\}=\frac{1}{(a+c)^2}\left( E\{X^2\}+2cE\{X\}+c^2 \right)=\frac{1}{(a+c)^2}(\sigma^2+\mu^2+2c\mu+c^2)=\frac{\sigma^2+(\mu+c)^2}{(a+c)^2}$$
Now we want the value of $c$ that minimizes $f(c)=\frac{\sigma^2+(\mu+c)^2}{(a+c)^2}, c\ge0$. We know the maximum or minimum of function is attained at $f'(c)=0$ or on the boundary $c=0$ or $c\rightarrow\infty$. Thus we have to calculate the solution to the equation $f'(c)=0$ to find our value.
$$f'(c)=\frac{2(\mu+c)(a+c)^2-2(a+c)(\sigma^2+(\mu+c)^2)}{(a+c)^4}=\frac{2(\mu+c)(a+c)-2(\sigma^2+(\mu+c)^2)}{(a+c)^3}=\frac{(a-\mu)c+(a\mu-\sigma^2)}{(a+c)^3}=0 \Rightarrow c=\frac{\sigma^2-a\mu}{a-\mu}$$
Now substituting $\mu=0$ gives $c_{\text{opt}}=\frac{\sigma^2}{a}$