Prove that $a^{4b}+b^{4a}\geq \frac{1}{2}$ Inspired by a problem of Vasile Cirtoaje I propose this :

Let $a,b>0$ such that $a+b=1$ then we have :
  $$a^{4b}+b^{4a}\geq \frac{1}{2}$$

I compute the derivative of $f(x)=x^{4(1-x)}+(1-x)^{4x}$ on $]0,1]$ we get :
$$ f'(x)=x^{4 (1 - x)} (\frac{4 (1 - x)}{x} - 4 \log(x)) + (1 - x)^{(4 x)} (4 \log(1 - x) - \frac{4 x}{1 - x})$$
If we denote by $g(x)$ the function :
$$g(x)=x^{4 (1 - x)} (\frac{4 (1 - x)}{x} - 4 \log(x))$$
We can rewrite the derivative as :
$$f'(x)=g(x)-g(1-x)$$
So it's remains to show that $g(x)\geq g(1-x)$ or $g(x)\leq g(1-x)$ 
So it remains to show that $g(x)$ is increasing or decreasing .
After that I'm stuck...
Any helps are very appreciated !
Thanks a lot for your time . 

Alternative proof:
I give a proof following @Malper's nice idea.

Fact 1: Let $x\in (0, 1/2]$ and $t \in (0, 1/2]$. Then $x^{4 - 4t} + (1 - x)^{4t} \ge \frac12$.
The proof is given at the end.
By Fact 1, we have
$x^{4 - 4x} + (1 - x)^{4x} \ge \frac12$
for all $x$ in $(0, 1/2]$.
We are done.

Proof of Fact 1: Let $f(x, t) = x^{4 - 4t} + (1 - x)^{4t}$. We have
\begin{align}
 \frac{\partial f}{\partial t}
 &= 4(1 - x)^{4t}\ln (1 - x) - 4x^{4 - 4t}\ln x\\
 &= 4x^{-4t}\Big([x(1 - x)]^{4t}\ln(1 - x) - x^4\ln x\Big)\\
 &\le 4x^{-4t}\Big([x(1 - x)]^{4\cdot \frac12}\ln(1 - x) - x^4\ln x\Big)\\
 &= 4x^{2 - 4t}\Big((1 - x)^2\ln(1 - x) - x^2\ln x\Big)\\
 &\le 0
\end{align}
where we have used $(1 - x)^2\ln(1 - x) - x^2\ln x \le 0$.
Note: Let $g(x) = (1 - x)^2\ln(1 - x) - x^2\ln x$. We have $g''(x) = 2\ln(1 - x) - 2\ln x \ge 0$ on $(0, 1/2]$. Thus, $g(x)$ is convex on $(0, 1/2]$. Also, $g(0^{+}) = 0$ and $g(1/2) = 0$. Thus, $g(x) \le 0$ on $(0, 1/2]$.
Thus, we have $f(x, t) \ge f(x, 1/2) = x^2 + (1 - x)^2 \ge 1/2$.
We are done.
