So here's one way of doing it, probably a little clumsy but trying to be as explicit as possible. Also note that I use both symmetry and antisymmetry brackets at the same time, so it's just about being careful with both. Stating with $$ \Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big( T^{a}{}_{bc} + T_{c}{}^{a}{}_b - T_{bc}{}^{a} \big) \ , $$ we can write this in terms of the connection (using antisymmetry brackets) as $$ \Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big(2\Gamma^{a}{}_{[bc]} +2 g_{cd} g^{ae} \Gamma^{d}{}_{[eb]} - 2 g_{bd}g^{ae} \Gamma^{d}{}_{[ce]} \big) \ . $$ Then symmetrising one finds $$ \begin{align} \Gamma^{a}{}_{(bc)} &= \big\{{}^a_{bc}\big\} - \frac{1}{2} \Big(0+2g_{(c|d} g^{ae}\Gamma^{d}{}_{[e|b)]}+2g_{(b|d} g^{ae}\Gamma^{d}{}_{[e|c)]} \Big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big( g_{(c|d}\Gamma^{d}{}_{e|b)}-g_{(c|d}\Gamma^{d}{}_{b)e}+g_{(b|d}\Gamma^{d}{}_{e|c)}+g_{(b|d}\Gamma^{d}{}_{c)e}\big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big(2g_{(c|d}\Gamma^{d}{}_{e|b)} -2g_{(c|d}\Gamma^{d}{}_{b)e}\big) \\ &= \big\{{}^a_{bc}\big\} + g^{ae} g_{(c|d}\Gamma^{d} {}_{b)e} - g^{ae}g_{(c|d}\Gamma^{d}{}_{e|b)} \\ &= \big\{{}^a_{bc}\big\} +g^{ae} g_{d(c} T^{d}{}_{b)e} \\ &= \big\{{}^a_{bc}\big\} + T_{(bc)}{}^{a} \ , \end{align} $$$$ \begin{align} \Gamma^{a}{}_{(bc)} &= \big\{{}^a_{bc}\big\} - \frac{1}{2} \Big(0+2g_{(c|d} g^{ae}\Gamma^{d}{}_{[e|b)]}+2g_{(b|d} g^{ae}\Gamma^{d}{}_{[e|c)]} \Big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big( g_{(c|d}\Gamma^{d}{}_{e|b)}-g_{(c|d}\Gamma^{d}{}_{b)e}+g_{(b|d}\Gamma^{d}{}_{e|c)}+g_{(b|d}\Gamma^{d}{}_{c)e}\big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big(2g_{(c|d}\Gamma^{d}{}_{e|b)} -2g_{(c|d}\Gamma^{d}{}_{b)e}\big) \\ &= \big\{{}^a_{bc}\big\} + g^{ae} g_{(c|d}\Gamma^{d} {}_{b)e} - g^{ae}g_{(c|d}\Gamma^{d}{}_{e|b)} \\ &= \big\{{}^a_{bc}\big\} +g^{ae} g_{d(c} T^{d}{}_{b)e} \\ &= \big\{{}^a_{bc}\big\} + T_{(bc)}{}^{a} \ \ , \end{align} $$ which gives the desired outcome. The first term on the first line vanishes from $[(ab)]=0$, and the Christoffel symbol of course stays the same. The last fewrest of the lines could probably be condensed into one too, but hopeare just manipulating indices whilst trying to stay clear and explicit. Hope this helps.
So here's one way of doing it, probably a little clumsy but trying to be as explicit as possible. Also note that I use both symmetry and antisymmetry brackets at the same time, so it's just about being careful with both. Stating with $$ \Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big( T^{a}{}_{bc} + T_{c}{}^{a}{}_b - T_{bc}{}^{a} \big) \ , $$ we can write this in terms of the connection (using antisymmetry brackets) as $$ \Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big(2\Gamma^{a}{}_{[bc]} +2 g_{cd} g^{ae} \Gamma^{d}{}_{[eb]} - 2 g_{bd}g^{ae} \Gamma^{d}{}_{[ce]} \big) \ . $$ Then symmetrising one finds $$ \begin{align} \Gamma^{a}{}_{(bc)} &= \big\{{}^a_{bc}\big\} - \frac{1}{2} \Big(0+2g_{(c|d} g^{ae}\Gamma^{d}{}_{[e|b)]}+2g_{(b|d} g^{ae}\Gamma^{d}{}_{[e|c)]} \Big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big( g_{(c|d}\Gamma^{d}{}_{e|b)}-g_{(c|d}\Gamma^{d}{}_{b)e}+g_{(b|d}\Gamma^{d}{}_{e|c)}+g_{(b|d}\Gamma^{d}{}_{c)e}\big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big(2g_{(c|d}\Gamma^{d}{}_{e|b)} -2g_{(c|d}\Gamma^{d}{}_{b)e}\big) \\ &= \big\{{}^a_{bc}\big\} + g^{ae} g_{(c|d}\Gamma^{d} {}_{b)e} - g^{ae}g_{(c|d}\Gamma^{d}{}_{e|b)} \\ &= \big\{{}^a_{bc}\big\} +g^{ae} g_{d(c} T^{d}{}_{b)e} \\ &= \big\{{}^a_{bc}\big\} + T_{(bc)}{}^{a} \ , \end{align} $$ which gives the desired outcome. The first term on the first line vanishes from $[(ab)]=0$, and the Christoffel symbol of course stays the same. The last few lines could probably be condensed into one too, but hope this helps.
So here's one way of doing it, probably a little clumsy but trying to be as explicit as possible. Also note that I use both symmetry and antisymmetry brackets at the same time, so it's just about being careful with both. Stating with $$ \Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big( T^{a}{}_{bc} + T_{c}{}^{a}{}_b - T_{bc}{}^{a} \big) \ , $$ we can write this in terms of the connection (using antisymmetry brackets) as $$ \Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big(2\Gamma^{a}{}_{[bc]} +2 g_{cd} g^{ae} \Gamma^{d}{}_{[eb]} - 2 g_{bd}g^{ae} \Gamma^{d}{}_{[ce]} \big) \ . $$ Then symmetrising one finds $$ \begin{align} \Gamma^{a}{}_{(bc)} &= \big\{{}^a_{bc}\big\} - \frac{1}{2} \Big(0+2g_{(c|d} g^{ae}\Gamma^{d}{}_{[e|b)]}+2g_{(b|d} g^{ae}\Gamma^{d}{}_{[e|c)]} \Big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big( g_{(c|d}\Gamma^{d}{}_{e|b)}-g_{(c|d}\Gamma^{d}{}_{b)e}+g_{(b|d}\Gamma^{d}{}_{e|c)}+g_{(b|d}\Gamma^{d}{}_{c)e}\big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big(2g_{(c|d}\Gamma^{d}{}_{e|b)} -2g_{(c|d}\Gamma^{d}{}_{b)e}\big) \\ &= \big\{{}^a_{bc}\big\} + g^{ae} g_{(c|d}\Gamma^{d} {}_{b)e} - g^{ae}g_{(c|d}\Gamma^{d}{}_{e|b)} \\ &= \big\{{}^a_{bc}\big\} +g^{ae} g_{d(c} T^{d}{}_{b)e} \\ &= \big\{{}^a_{bc}\big\} + T_{(bc)}{}^{a} \ \ , \end{align} $$ which gives the desired outcome. The first term on the first line vanishes from $[(ab)]=0$, and the Christoffel symbol of course stays the same. The rest of the lines are just manipulating indices whilst trying to stay clear and explicit. Hope this helps.
So here's one way of doing it, probably a little clumsy but trying to be as explicit as possible. Also note that I use both symmetry and antisymmetry brackets at the same time, so it's just about being careful with both. Stating with $$ \Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big( T^{a}{}_{bc} + T_{c}{}^{a}{}_b - T_{bc}{}^{a} \big) \ , $$ we can write this in terms of the connection (using antisymmetry brackets) as $$ \Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big(2\Gamma^{a}{}_{[bc]} +2 g_{cd} g^{ae} \Gamma^{d}{}_{[eb]} - 2 g_{bd}g^{ae} \Gamma^{d}{}_{[ce]} \big) \ . $$ Then symmetrising one finds $$ \begin{align} \Gamma^{a}{}_{(bc)} &= \big\{{}^a_{bc}\big\} - \frac{1}{2} \Big(0+2g_{(c|d} g^{ae}\Gamma^{d}{}_{[e|b)]}+2g_{(b|d} g^{ae}\Gamma^{d}{}_{[e|c)]} \Big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big( g_{(c|d}\Gamma^{d}{}_{e|b)}-g_{(c|d}\Gamma^{d}{}_{b)e}+g_{(b|d}\Gamma^{d}{}_{e|c)}+g_{(b|d}\Gamma^{d}{}_{c)e}\big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big(2g_{(c|d}\Gamma^{d}{}_{e|b)} -2g_{(c|d}\Gamma^{d}{}_{b)e}\big) \\ &= \big\{{}^a_{bc}\big\} + g^{ae} g_{(c|d}\Gamma^{d} {}_{b)e} - g^{ae}g_{(c|d}\Gamma^{d}{}_{e|b)} \\ &= \big\{{}^a_{bc}\big\} +g^{ae} g_{d(c} T^{d}{}_{b)e} \\ &= \big\{{}^a_{bc}\big\} + T_{(bc)}{}^{a} \ , \end{align} $$ which gives the desired outcome. The first term on the first line vanishes from $[(ab)]=0$, and the Christoffel symbol of course stays the same. The last few lines could probably be condensed into one too, but hope this helps.