Can the relation a^p+b^p>=2c^p be proven for p values between 0 and 1?

[onako]

Given $$a+b >=c$$, I'm trying to obtain values of p such that $$a^p+b^p>=c^p$$. Obviously,
for p>1 the relation does not work. However, I wonder if it can be proved that it works for p from the interval [0, 1].
I tried to solve this with the nth power of binomial a+b, but the formula I found is complicated. Perhaps you
have a much simpler approach to resolve the problem (either by disproving that the requirement is possible, or by proving the opposite)

Thanks

 

[onako]

So, to state the problem precisely:
Assuming $$a+b>=2c$$ and $$a^p*b^p >= c^{2p}, p \in [0, 1]$$
can we prove that
$$a^p+b^p>=2c^p$$
This holds for for p=0.5, but I wonder if this applies to the other values
0<p<1. Perhaps it is true for 0.5<=p<=1.

Thanks, I hope this clarified the problem. Please move the post to the appropriate math subforum.