<s>[INST] $x^2+xy+xz+yz=6+2\sqrt{5}$. Find the minimum of $3x+y+2z$ Let $x,~y,~z>0$ satisfy $x^2+xy+xz+yz=6+2\sqrt{5}$. How to find the minimum of $3x+y+2z$? I can find it by Larange multiplier in calculus, but I wonder if there is a easy way using, say Cauchy-Schwartz inequality?
 [/INST] You're given $(x+y)(z+x)=6+2\sqrt{5}\iff (x+y)(2z+2x)=12+4\sqrt{5}$. Thus with AM-GM $$3x+y+2z=(x+y)+(2z+2x)\geqslant 2\cdot \sqrt{12+4\sqrt{5}}$$
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