Little result using RCF theorem and an inequality by Vasile Cirtoaje

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Well , I want to share with you a little result :

The function $$f(x)=(x)^{2(1-x)}$$ is concave on $I=[\frac{1}{2},1]$

Moreover Vasile Cirtoaje proved that : $$f(x)+f(1-x)\leq 1\tag{1}$$

So we can apply the RCF theorem since $1-x+x=2\cdot\frac{1}{2}$ and with $(1)$ it implies that :

Let $1>x_1>0,1>x_2>0$ be two real numbers such that $x_1+x_2\geq 1$ then we have : $$f(x_1)+f(x_2)\leq 2 f\Big(\frac{x_1+x_2}{2}\Big)\quad (E)$$

Well my question :

Have you an alternative proof ?

Thanks in advance !

Update:

If $x_1,x_2\in[0,5,1]$ then we can directly apply Jensen's inequality .Well it's a first important remark . For the second case $x_1,x_2\in(0,1]$ we can perhaps use Bernoulli's inequality , take logarithm on both side and then use derivatives .If you have a nice idea make a comments/answers . I recall that my question concern the proof of $(E)$ .

Reference : 1 Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions", The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137. https://eudml.org/doc/223938