Positivity of power function. Prove that 
           $6^a-7^a+2\cdot 4^a-3^a-5^a\ge0$ for $-\frac{1}{2}\le a\le0$.
I tried to do it by first derivative test but derivative become more complicated (same with 2nd derivative for convexity). 
Here is it's plotting.

We want to show that 
$$\left((3^x+7^x+5^x)/3\right)^{1/x} > \left((6^x+4^x+4^x)/3\right)^{1/x}\hbox{ for }-0.5\le x<0$$ "as it may be seen from the plot".
It's not too hard to test that 
$$\hbox{(1) }4.57 < \left((3^{-\frac{1}{2}}+7^{-\frac{1}{2}}+5^{-\frac{1}{2}})/3\right)^{-2} $$
$$\hbox{(2) }\left((6^{-\frac{1}{4}}+4^{-\frac{1}{4}}+4^{-\frac{1}{4}})/3\right)^{-4} < 4.57$$
$$\hbox{(3) }\sqrt[3]{6\cdot 4\cdot 4} < \left((3^{-\frac{1}{4}}+7^{-\frac{1}{4}}+5^{-\frac{1}{4}})/3\right)^{-4}$$
So, using Generalized mean inequality we have
$$M_x(6,4,4)< M_{-\frac{1}{4}}(6,4,4)<M_{-\frac{1}{4}}(4.57,4.57,4.57)=$$
$$M_{-\frac{1}{2}}(4.57,4.57,4.57)<M_{-\frac{1}{2}}(3,5,7)\le M_x(3,5,7)\ \forall x:-\frac{1}{2}\le x<-\frac{1}{4},$$
$$M_x(6,4,4)\le M_0(6,4,4)<M_{-\frac{1}{4}}(3,5,7)<M_x(3,5,7)\ \forall x:-\frac{1}{4}\le x<0.$$
