<s>[INST] 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.
 [/INST] 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 $-\frac12\le a<0$.
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