How to prove this $\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$ Let $x,y>0$, and $x+y=2$, show that
$$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$$
I tried using Minkowski inequality
$$\sqrt{x^2+3}+\sqrt{y^2+3}\ge\sqrt{(x+y)^2+(\sqrt{3}+\sqrt{3})^2}=\sqrt{4+12}=4$$
But
$$\sqrt{xy+3}\le\sqrt{\dfrac{(x+y)^2}{4}+3}=2$$
so i'm not sure how it would work with this problem,Thanks
 A: To prove that
$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$,
We use Jensen's inequality.
$1*\sqrt{x^2+3}+1*\sqrt{y^2+3}+1*\sqrt{xy+3}\ge (1+1+1) \sqrt{\frac{x^2+y^2+xy+9}
{1+1+1}}$
Then, since $x+y=2$, we have:
$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3} \ge 3\sqrt{\frac{x^2+(2-x)^2+x(2-x)+9}{3}}$
$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3} \ge 3\sqrt{\frac{x^2-2x+13}{3}} = 3\sqrt{\frac{(x-1)^2+12}{3}} \ge 6$
A: Let $\sqrt{x^2+3}+\sqrt{y^2+3}=t$.
We need to prove that
$$\sqrt{x^2+3}+\sqrt{y^2+3}-4\geq2-\sqrt{xy+3}$$ or
$$\frac{x^2+3+y^2+3+2\sqrt{(x^2+3)(y^2+3)}-16}{t+4}\geq\frac{1-xy}{2+\sqrt{xy+3}}$$ or
$$\frac{2(x^2+3)+2(y^2+3)-16-\left(\sqrt{x^2+3}-\sqrt{y^2+3}\right)^2}{t+4}\geq\frac{4-4xy}{4(2+\sqrt{xy+3})}$$ or
$$\frac{(x-y)^2-\frac{(x-y)^2(x+y)^2}{t^2}}{t+4}\geq\frac{(x-y)^2}{4(2+\sqrt{xy+3})}$$ or
$$\frac{4(t^2-4)(2+\sqrt{xy+3})}{t^2(t+4)}\geq1$$
and since $xy\geq0$, it's enough to prove that $f(t)\geq\frac{2-\sqrt3}{4}$ , where $f(t)=\frac{t^2-4}{t^2(t+4)}$.
But by Mikowski $t=\sqrt{x^2+3}+\sqrt{y^2+3}\geq\sqrt{(x+y)^2+3(1+1)^2}=4$.
In another hand, 
$$t=\sqrt{x^2+y^2+6+2\sqrt{(x^2+3)(y^2+3)}}=\sqrt{10-2xy+2\sqrt{x^2y^2+3(x^2+y^2)+9}}=$$
$$=\sqrt{10-2xy+2\sqrt{x^2y^2-6xy+21}}\leq\sqrt{10-2xy+2(xy+\sqrt{21})}=\sqrt3+\sqrt7.$$
Also we have
$$f'(t)=\frac{-t^3+12t+32}{t^3(t+4)^2}>0$$
for all $t\in\left[4,\sqrt3+\sqrt7\right)$.
Thus, $f(t)\geq f(4)=\frac{3}{32}>\frac{1}{4}(2-\sqrt3)$ and we are done!
