Let $a,b,c$ the sides of a triangle such that $a+b+c= \dfrac{1}{34^4}$. Show that: $$ \sqrt[5]{a+b-c} + 16\sqrt[5]{a+c-b} + \sqrt[5]{b+c-a} \leq 1$$
I put $a = x+y, b=y+z$ and $c=z+x$ with $(x,y,z) \in \mathbb{R}_{+}^3$. So I have to prove that $$ \sqrt[5]{2y} + 16\sqrt[5]{2x} + \sqrt[5]{2z} \leq 1 $$ and I think that I should maybe use minkowsky inequality but I can't make it in a good way. Thanks for your help!
By Holder(https://math.stackexchange.com/tags/holder-inequality/info) we obtain: $$\sqrt[5]{a+b-c}+16\sqrt[5]{a+c-b}+\sqrt[5]{b+c-a}\leq$$ $$\leq\sqrt[5]{\left(1+32+1\right)^4(a+b-c+a+c-b+b+c-a)}=1.$$