The Many Faces Of An Inequality

Let $a=x$ , $b=2y$ , and $c=3z$ . We have $a+b+c=2\sqrt3$ . The expression then becomes $\sum\limits_{a,b,c} \sqrt{a^2+ab+b^2}$ . Note that $a^2+ab+b^2=\dfrac{1}{2}\left(a^2+b^2+(a+b)^2\right)$ . By Cauchy-Schwarz Inequality, $(a^2+b^2)\geq\dfrac{(a+b)^2}{2}$ . Then $\sum\limits_{a,b,c} \sqrt{a^2+ab+b^2}=\sum\limits_{a,b,c} \sqrt{\dfrac{1}{2} \left(a^2+b^2+(a+b)^2\right) }\geq \sum\limits_{a,b,c} \sqrt{\dfrac{1}{2} \left(\dfrac{(a+b)^2}{2}+(a+b)^2\right)}=\sum\limits_{a,b,c} \sqrt{\dfrac{3}{4} (a+b)^2}=\dfrac{\sqrt3}{2}\left[(a+b)+(b+c)+(c+a)\right] = \dfrac{\sqrt3}{2}(2)(2\sqrt3)=6$

We can write the expression above as $\sqrt{(x+y)^2+(y\sqrt{3})^2}+\sqrt{(2y+\frac {3z}{2})^2+(\frac {3z\sqrt{3}}{2})^2}+\sqrt{(3z+\frac {x}{2})^2+(\frac {x\sqrt{3}}{2})^2}$ By minkowski inequality the minimum value will be $\sqrt{(\frac{3x}{2}+3y+\frac{9z}{2})^2+(\frac{x\sqrt3}{2}+y\sqrt{3}+\frac {3z\sqrt{3}}{2})^2}$ = $\sqrt{(\frac{3}{2})^2(x+2y+3z)^2+(\frac{\sqrt3}{2})^2(x+2y+3z)^2}$ = $\sqrt{27+9}$ = $6$

Rewrite the expression $A=\sqrt{(x+y)^2+3y^2}+\sqrt{(y+3z)^2+3y^2}+\sqrt{\big(x+\frac{3}{2}z\big)^2+\frac{27}{4}z^2}$ .By Cauchy-Schwarz we have $\sqrt{\frac{4}{3}\big[(x+y)^2+3y^2\big]}\geq |x+2y|$ $\sqrt{\frac{4}{3}\big[(y+3z)^2+3y^2\big]}\geq |2y+3z|$ $\sqrt{\frac{4}{3}\bigg[\big(x+\frac{3}{2}z\big)^2+\frac{27}{4}z^2\bigg]}\geq |x+3z|$ So $\frac{2\sqrt{3}}{3}A\geq |x+2y|+|2y+3z|+|x+3z|\geq |2(x+2y+3z)|=4\sqrt{3}$ . Therefore $A\geq 6$ , the equality holds when $(x,y,z)=\bigg(\frac{2\sqrt{3}}{3};\frac{\sqrt{3}}{3};\frac{2\sqrt{3}}{9}\bigg)$