<s>[INST] How to prove $\sqrt{x(3x+y)} + \sqrt{y(3y+z)} +\sqrt{z(3z+x)} \le 2(x+y+z)$ 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
 [/INST] 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).
$$2(x+y+z)=\frac{1}{4}\sum_{cyc}(4x+3x+y)=\frac{1}{4}\sum_{cyc}\left(\sqrt{(4x)^2+0^2}+\sqrt{(3x+y)^2+0^2}\right)\geq$$
$$\geq\frac{1}{4}\sum_{cyc}\sqrt{(4x+3x+y)^2+(0+0)^2}=\frac{1}{4}\sum_{cyc}\sqrt{49x^2+14xy+y^2}=$$
$$=\frac{1}{4}\sum_{cyc}\sqrt{48x^2+16xy+(x-y)^2}\geq\frac{1}{4}\sum_{cyc}\sqrt{48x^2+16xy}=\sum_{cyc}\sqrt{x(3x+y)}.$$
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