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
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:
$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3} \ge 3\sqrt{\frac{x^2+(2-x)^2+x(2-x)+9}{3}}$
$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3} \ge 3\sqrt{\frac{x^2-2x+13}{3}} = 3\sqrt{\frac{(x-1)^2+12}{3}} \ge 6$
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!
It is easy to prove that $$\sqrt{xy+3} = \sqrt{5 - (x^2+y^2)/2} \ge \frac{16}{7} - \frac17(x^2 + y^2).$$ (Note: Use $x^2 + y^2 \in [2, 4]$. Also, see the comment for this answer.)
It suffices to prove that
$$\sqrt{x^2 + 3} + \sqrt{y^2 + 3} + \frac{16}{7} - \frac17(x^2 + y^2) \ge 6,$$
or
$$\sqrt{x^2 + 3} + \sqrt{y^2 + 3} + \frac{16}{7} - \frac17(x^2 + y^2) - \frac{3}{14}(x + y - 2) \ge 6,$$
which is written as
$$\left(\sqrt{x^2 + 3} - \frac{23}{14} - \frac17x^2 - \frac{3}{14}x \right)
+ \left(\sqrt{y^2 + 3} - \frac{23}{14} - \frac17y^2 - \frac{3}{14}y \right) \ge 0. \tag{1}$$
(1) is true since
$\sqrt{u^2 + 3} - \frac{23}{14} - \frac17u^2 - \frac{3}{14}u \ge 0$
for all $u\in [0, 2]$.
(Note: $(u^2 + 3) - (\frac{23}{14} + \frac17u^2 + \frac{3}{14}u)^2 =
\frac{1}{196}(59 - 20u - 4u^2)(u-1)^2 \ge 0$.)
We are done.
