Show $2^{10x/7}(1-x) \le \frac12$ for sufficiently small $x>0$ 
Show  $2^{10x/7}(1-x) \le \frac12$ for sufficiently small $x$, where $x$ is positive.

I've thought about trying to find the limit to this as $x$ tended to $0$, but have had a lot of difficulties. Help would be greatly appreciated.
 A: As Barry Cipra commented, there is something wrong in the problem.
Considering the function $$f(x)=2^{10 x/7} (1-x)$$ $$f'(x)=-\frac{1}{7} 2^{10 x/7} (10 (x-1) \log (2)+7)$$ the first derivative cancels for $x_*=1-\frac{7}{\log (1024)}\approx -0.00988653$ for which $f(x_*)=\frac{7\ 2^{3/7}}{e \log (32)}\approx 1.00005$ and the second derivative test shows that this is a maximum.
Close to $x=0$, a Taylor expansion gives $$f(x)=1+ \left(\frac{10 \log (2)}{7}-1\right)x+ \left(\frac{50 \log ^2(2)}{49}-\frac{10
   \log (2)}{7}\right)x^2+O\left(x^3\right)$$ Close to $x=1$, a Taylor expansion gives $$f(x)= 2^{10/7} (1-x)-\frac{20}{7}  \left(2^{3/7} \log
   (2)\right)(1-x)^2+O\left((x-1)^3\right)$$
In order to have $f(x)\lt \frac 12$, it would be required that $$x>1+\frac{7 }{\log (1024)}W\left(-\frac{5 \log (2)}{14\ 2^{3/7}}\right)\approx 0.76576$$ So, I suppose that the problem is for values of $x$ close to $1$ instead.
A: to make it clear, the value of $y=2^{(10x/7)}(1-x)$ between $[0,1]$ looks like:

obviously it's not less than 1/2 for small positive x 
by the way, x=0 is where $2^{(10x/7)}(1-x)$ get max value
two roots for $2^{(10x/7)}(1-x)=1/2$ is approximately -1.7049 and 0.76576
