For $a,b,c\in \mathbb{R}_{+}$ and $1<p<2$ does the following estimate hold:
$$(a+b+c)^{p-1} \leq C (a^{p-1} + b^{p-1} + c^{p-1})$$
for a large enough constant?
I tried to prove this in the following way:
$(a+b)^{p-1}\leq 2^{p-1}(a^{p-1}+ b^{p-1})$
and so
$$(a+b+c)^{p-1} \leq 2^{p-1} ( (a+b)^{p-1} + c^{p-1}) \leq 2^{2(p-1)}(a^{p-1}+b^{p-1} + c^{p-1}).$$
Is this reasoning correct?
You can get an even better estimate: Consider $g: [0, \infty)^2 \rightarrow \mathbb{R}$ where $$ g(x, y) := (x+y)^{p-1}-x^{p-1}-y^{p-1}. $$ Note, since $p-2<0$: $$ \partial_x g(x, y) = (p-1)((x+y)^{p-2} - x^{p-2}) \leq (p-1)(x^{p-2} - x^{p-2}) = 0 $$ You can show the same way that $\partial_y g(x, y) \leq 0$. So $g$ is decreasing in $x$ and $y$ and $g(0, 0) = 0$. Therefore $g$ is nonpositive everywhere. It follows that $$ (x+y)^{p-1} \leq x^{p-1} + y^{p-1}. $$ You can iterate this argument to receive $$ (a+b+c)^{p-1} \leq a^{p-1} + b^{p-1} + c^{p-1}. $$
