# If $a,b >0.$Prove:$\frac{\ln⁡(Γ(2a+1))}{a}+\frac{\ln⁡(Γ(2b+1))}{b}\leq \frac{4\ln⁡(Γ(a+b+1))}{(a+b)}$

If $$a,b >0.Prove:\frac{\ln⁡(Γ(2a+1))}{a}+\frac{\ln⁡(Γ(2b+1))}{b}\leq \frac{4\ln⁡(Γ(a+b+1))}{(a+b)}$$

I think it is rlated to convexity of Gamma function. I tried to se convexity of Gamma and concavity og log.

Question is proposed by Jalil Hajimir to Romanian Math Magazine.