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.
 A: For  $a\in[-\frac12,0)$ and $x>0$ let $$g_a(x)=x^a-(x+1)^a-(x+3)^a$$
Then  $$f(a)=g_a(4)-g_a(3)$$
By the Mean Value Theorem this equals $g_a'(\theta)=ag_{a-1}(\theta)$ for some $\theta\in [3,4]$.  We have
$$\begin{align}\frac{g_{a-1}(\theta)}{\theta^{a-1}}&=1-(1+\tfrac1\theta)^{a-1}-(1+\tfrac3\theta)^{a-1}\\&\le 1-(1+\tfrac13)^{a-1}-2^{a-1}&\text{(negative exponent $\Rightarrow$ maximize $\tfrac1\theta$)}\\
&\le 1-(\tfrac43)^{-3/2}-2^{-3/2}&\text{(bases $>1$ $\Rightarrow$ minimize neg. exponent)}\\
&=\tfrac18\left(8-3\sqrt3-2\sqrt 2\right).\end{align} $$
We have $5<3\sqrt 3<6$ and $2<1\sqrt 2<3$, therefore the sign of the expression in parenthesis is not immediately clear (without using a calculator). But the conjugates $8+3\sqrt3-2\sqrt 2$, $8-3\sqrt3+2\sqrt 2$, $8+3\sqrt3+2\sqrt 2$ are all definitely positive (in fact $>8+2-6=4$).
The product of all four conjugates however is
$$\begin{align}&(8-3\sqrt3-2\sqrt 2)(8-3\sqrt3+2\sqrt 2)(8+3\sqrt3-2\sqrt 2)(8+3\sqrt3+2\sqrt 2)\\=\:&\bigl((8-3\sqrt 3)^2-8\bigr)\bigl((8+3\sqrt 3)^2-8\bigr)\\=\:&(83-48\sqrt 3)(83+48\sqrt 3)\\=\:&83^2-3\cdot 48^2\\
=\:&{-23}.\end{align} $$
We conclude $8-3\sqrt 3-2\sqrt 2<0$, so that
$$f(a)=ag_{a-1}(\theta)\ge a\theta^{a-1}\tfrac18\left(8-3\sqrt3-2\sqrt 2\right)>0 $$
for $-\frac12\le a<0$.
A: 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
$$M_x(6,4,4)< M_{-\frac{1}{4}}(6,4,4)<M_{-\frac{1}{4}}(4.57,4.57,4.57)=$$
$$M_{-\frac{1}{2}}(4.57,4.57,4.57)<M_{-\frac{1}{2}}(3,5,7)\le M_x(3,5,7)\ \forall x:-\frac{1}{2}\le x<-\frac{1}{4},$$
$$M_x(6,4,4)\le M_0(6,4,4)<M_{-\frac{1}{4}}(3,5,7)<M_x(3,5,7)\ \forall x:-\frac{1}{4}\le x<0.$$
A: $f(a) = 6^a - 7^a + 2 \cdot 4^a - 3^a - 5^a$.


*

*Value of function


$f(-1/2) = 0.00572$, and $f(0) = 0$.


*

*First derivative:


$2^{2 + 2 a} Log[2] - 3^a Log[3] - 5^a Log[5] + 6^a Log[6] - 7^a Log[7]$.
Evaluated at $a = -1/2$ and $0$:
${d f(a) \over d a} |_{a=-1/2} = 0.02825$ and ${d f(a) \over d a} |_{a=0} = -0.0896$.


*

*Second derivative:


${d^2 f(a) \over d a} = 2^{2 a+3} \log ^2(2)-3^a \log ^2(3)-5^a \log ^2(5)+6^a \log ^2(6)-7^a \log ^2(7)$
which is always negative in the range $-1/2 \leq a \le 0$.  We know this because the second derivative is negative at each end of the range and the third derivative is always negative (as can be seen on a plot).
QED.
