Prove $\sqrt{x(3x+y)} + \sqrt{y(3y+z)} +\sqrt{z(3z+x)} \le 2(x+y+z)$ using triangle inequality
To answer the comment: I'm pretty stumped on the question, i know i can substitute some variables in to make it simpler but from there im clueless, thats why im here, this is my last resort haha
By AM-GM $$\sum_{cyc}\sqrt{x(3x+y)}=\sqrt{\sum_{cyc}(3x^2+xy+2\sqrt{xy(3x+y)(3y+z)}}=$$ $$=\sqrt{\sum_{cyc}\left(3x^2+xy+2\sqrt{x(3y+z)\cdot y(3x+y)}\right)}\leq$$ $$\leq\sqrt{\sum_{cyc}(3x^2+xy+x(3y+z)+y(3x+y))}=2(x+y+z).$$ Done!
The proof by triangle inequality (Minkowski). $$2(x+y+z)=\frac{1}{4}\sum_{cyc}(4x+3x+y)=\frac{1}{4}\sum_{cyc}\left(\sqrt{(4x)^2+0^2}+\sqrt{(3x+y)^2+0^2}\right)\geq$$ $$\geq\frac{1}{4}\sum_{cyc}\sqrt{(4x+3x+y)^2+(0+0)^2}=\frac{1}{4}\sum_{cyc}\sqrt{49x^2+14xy+y^2}=$$ $$=\frac{1}{4}\sum_{cyc}\sqrt{48x^2+16xy+(x-y)^2}\geq\frac{1}{4}\sum_{cyc}\sqrt{48x^2+16xy}=\sum_{cyc}\sqrt{x(3x+y)}.$$