min. of log expression

Hi,I have a solution, I think it is correct...Check it and do tell me...

c\geq 2 \implies c-1 \geq \frac c 2 \geq \frac c b

\implies bc\geq b+ c, \implies \ln b + \ln c \geq \ln(b+c)

\implies \frac{1}{\ln(b+c)}\geq \frac{1}{\ln b + \ln c}

\implies \log _{b+c}a= \frac{\ln a}{\ln(b+c)}\geq \frac{\ln a}{\ln b + \ln c}

\implies \sum_{cyc}^{{}}\log _{b+c}a= \sum_{cyc}^{{}}\frac{\ln a}{\ln(b+c)}\geq \sum_{cyc}^{{}}\frac{\ln a}{\ln b + \ln c}\geq \frac{3}{2}

The last step being true by Nesbitt's Inequality...

Equality holds when a=b=c=2

lnx is concave which gives ln(b+c/2)≥ (lna + lnb)/2

so i dont think we can....??

hmm..