Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that$$\color{black}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{2(ab+bc+ca)+12}.}$$ Equality holds at $a=b=c=1$ or $a=b=0;c=3.$
I tried to use the following inequality$$\color{black}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \frac{9+ab+bc+ca}{2\sqrt{2}}.}$$ But it doesn't help since $$\sqrt{2(ab+bc+ca)+12}\ge \frac{9+ab+bc+ca}{2\sqrt{2}}.$$
Hope you can help me solve this problem. Thank you.
Fact. If $x,y,z\ge 0$ such that$$x^2y^2+y^2z^2+z^2x^2+\frac{33}{16}x^2y^2z^2\ge \frac{81}{16},$$then $x+y+z\ge 3.$
I proved the fact here.
Let $a+b+c=p=3; 0\le q=ab+bc+ca\le 3; r=abc.$ Now, we apply the fact where $$x=\sqrt{\frac{9(a+b)}{2(ab+bc+ca)+12}};y=\sqrt{\frac{9(c+b)}{2(ab+bc+ca)+12}};z=\sqrt{\frac{9(a+c)}{2(ab+bc+ca)+12}}.$$ Thus, it's enough to prove $$x^2y^2+y^2z^2+z^2x^2+\frac{33}{16}x^2y^2z^2\ge \frac{81}{16}.$$ It turns out $$\frac{81}{(2(ab+bc+ca)+12)^2}\cdot\sum_{cyc}(a+b)(a+c)+\frac{33}{16}\cdot\frac{729(a+b)(b+c)(c+a)}{(2(ab+bc+ca)+12)^3},$$
or $$\frac{81}{4(q+6)^2}\cdot(q+9)+\frac{33}{16}\cdot\frac{729}{8(q+6)^3}\cdot(3q-r)\ge \frac{81}{16}.$$ It can be rewritten as $$32(q^2+15q+54)+297(3q-r)\ge 8(q+6)^3. \tag{1}$$ We use the following result $$r \le \dfrac{q^2(p^2-q)}{2p(2p^2-3q)}=\dfrac{q^2(q-9)}{18(q-6)}.\tag{2}$$ (Proof of $(2)$ is below.)
From $(1)$ and $(2),$ we need to prove $$32(q^2+15q+54)+297(3q-\dfrac{q^2(q-9)}{18(q-6)})\ge 8(q+6)^3,$$which is equivalent to $$\frac{q(3-q)[ 16q(3-q) +257(3-q)+1257]}{6-q} \ge 0.$$The last inequality is obviously true for $q\in[0;3].$
Hence, the proof is done.
Proof of $(2).$
Notice that $$(a-b)^2(b-c)^2(c-a)^2 \ge 0,$$or$$p^2q^2-4q^3+2p(9q-2p^2)r-27r^2 \ge 0.$$ It is a quadratic of $r$ and$$\dfrac{p(9q-2p^2)-2\sqrt{(p^2-3q)^3}}{27} \le r \le \dfrac{p(9q-2p^2)+2\sqrt{(p^2-3q)^3}}{27}.$$ By using AM-GM$$27r \le p(9q-2p^2)+2\sqrt{(p^2-3q)^3}$$ \begin{align*} &=p(9q-2p^2)+\dfrac{(p^2-3q).2.\left(p^2-\dfrac{3}{2}q\right).p\sqrt{p^2-3q}}{p\left(p^2-\dfrac{3}{2}q\right)}\\ & \le p(9q-2p^2)+\dfrac{(p^2-3q)\left[\left(p^2-\dfrac{3}{2}q\right)^2+p^2(p^2-3q)\right]}{p\left(p^2-\dfrac{3}{2}q\right)}\\ &=\dfrac{27q^2(p^2-q)}{2p(2p^2-3q)} \end{align*} or $$r \le \dfrac{q^2(p^2-q)}{2p(2p^2-3q)}.$$
Proof of the fact.
Assuming that $0<x+y+z<3,$ let $m=kx;n=ky;p=kz$ where $k=\dfrac{3}{x+y+z}>1.$
Now, $m>x,n>y,p>z$ and $m+n+p=3.$ We'll prove$$m^2n^2+n^2p^2+p^2m^2+\frac{33}{16}m^2m^2p^2\le \frac{81}{16} (*).$$ Let $p=m+n+p; q=mn+np+pm; r=mnp.$ Rewrite $(*)$ in form $pqr$, a function of $r$$$f(r)=-\frac{33}{16}r^2+6r+\frac{81}{16}-q^2\ge 0,$$where $f'(r)=6-\dfrac{33}{8}r>0$ thus $f(r)$ is increasing.
By Schur of third degree, \begin{align*} f(r)&\ge f\left(\frac{4q-9}{3}\right)\\&=\frac{-33}{16}\left(\frac{4q-9}{3}\right)^2+6\left(\frac{4q-9}{3}\right)+\frac{81}{16}-q^2\\&=\frac{14}{3}(3-q)\left(q-\frac{9}{4}\right)\ge 0,\forall \frac{9}{4}\le q\le 3. \end{align*} In case $0\le q\le \dfrac{9}{4}:$ $f(r)=3r\left(2-\dfrac{11}{16}r\right)+\left(q+\dfrac{9}{4}\right)\left(-q+\dfrac{9}{4}\right)\ge 0.$
Hence, the $(*)$ is proven.
Also by the condition, $m>x,n>y,p>z$ $$m^2n^2+n^2p^2+p^2m^2+\frac{33}{16}m^2m^2p^2>x^2y^2+y^2z^2+z^2x^2+\frac{33}{16}x^2y^2z^2\ge \frac{81}{16} \tag{**}.$$
From $(*)$ and $(**)$ we get a contradiction.
In conclusion, the fact is proven.
Let $a^2+b^2+c^2=x(ab+ac+bc).$
Thus, $x\geq1$ and by C-S and Muirhead we obtain: $$\sum_{cyc}\sqrt{a+b}=\sqrt{2(a+b+c)+2\sqrt{(a+b)(a+c)}}=$$ $$=\sqrt{6+2\sqrt{\sum_{cyc}\left(a^2+3ab+2(a+b)\sqrt{(a+c)(b+c)}\right)}}\geq$$ $$\geq\sqrt{6+2\sqrt{\sum_{cyc}\left((a^2+3ab+2(a+b)(\sqrt{ab}+c)\right)}}\geq\sqrt{6+2\sqrt{\sum_{cyc}(a^2+11ab)}}$$ and it's enough to prove that: $$(a+b+c)\sqrt{(a^2+11ab)}\geq\sum_{cyc}(a^2+5ab)$$ or $$\sqrt{(x+2)(x+11)}\geq x+5$$ or $$x\geq1$$ and we are done!
Here is a proof using the isolated fudging.
After squaring both sides, it suffices to prove that $$\sum_{\mathrm{cyc}} 2\sqrt{(a + b)(b + c)} \ge 2(ab + bc + ca) + 6$$ or $$\sum_{\mathrm{cyc}} \left(2\sqrt{(a + b)(b + c)} - \frac{6b + b(a + c) + 4ca}{3} \right)\ge 0.$$
It suffices to prove that $$2\sqrt{(a + b)(b + c)} - \frac{6b + b(a + c) + 4ca}{3} \ge 0$$ or $$4(a + b)(b + c) - \left(\frac{6b + b(a + c) + 4ca}{3}\right)^2 \ge 0$$ or (using $b = 3 - a - c$) $$\frac19(a - c)^2\Big(6( 3 - a - c) + 9 - (a - c)^2\Big) \ge 0$$ which is true.
We are done.
