Let $a,b,c>0$. Prove that: $$\frac{\sqrt{a+b}}{c+\sqrt{ab}}+\frac{\sqrt{b+c}}{a+\sqrt{bc}}+\frac{\sqrt{c+a}}{b+\sqrt{ca}}\ge \frac{3\sqrt{2}}{2}\sqrt{\frac{a+b+c}{ab+bc+ca}}.$$
I used AM-GM: $\sqrt{ab}\le \frac{a+b}{2}$ and we will prove $$\sum_{cyc}\frac{\sqrt{a+b}}{c+\sqrt{ab}}\ge 2\sum_{cyc}\frac{\sqrt{a+b}}{2c+a+b},$$but $$\sum_{cyc}\frac{\sqrt{a+b}}{2c+a+b}\ge \frac{3\sqrt{2}}{2}\sqrt{\frac{a+b+c}{ab+bc+ca}}$$is wrong when $a=b=1;c=\dfrac{1}{5}$. This counter example also show that the wrong inequality doesn't help$$3\sqrt[3]{\frac{\sqrt{a+b}}{c+\sqrt{ab}}.\frac{\sqrt{b+c}}{a+\sqrt{bc}}.\frac{\sqrt{c+a}}{b+\sqrt{ca}}}\ge \frac{3\sqrt{2}}{2}\sqrt{\frac{a+b+c}{ab+bc+ca}}.$$ Again, we think of Holder inequality$$\left(\sum_{cyc}\frac{\sqrt{a+b}}{c+\sqrt{ab}}\right)^2.\sum{cyc}(a+b)^2(c+\sqrt{ab})^2\ge 8(a+b+c)^3,$$which implies that it suffice to prove $$(a+b+c)(ab+bc+ca)\ge \frac{3}{16}\sum{cyc}(a+b)^2(c+\sqrt{ab})^2.$$ I think this inequality is true and the equality holds at $(t,t,t); (0,0,t)$. Am I in right way ?
Yes, your Holder gives a proof, but it should be: $$(a+b+c)^2(ab+ac+bc)\geq\frac{9}{16}\sum_{cyc}(a+b)^2(c+\sqrt{ab})^2,$$ which is true.
Indeed, we need to prove that: $$16(a^2+b^2+c^2)^2(a^2b^2+a^2c^2+b^2c^2)\geq9\sum_{cyc}(a^2+b^2)^2(c^2+ab)^2$$ for non-negatives $a$, $b$ and $c$,which we can kill by $uvw$: https://artofproblemsolving.com/community/c6h278791
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that: $$16(9u^2-6v^2)^2(9v^4-6uw^3)\geq9(-9u^2w^6+84v^2w^6-486u^5w^3+810u^3v^2w^3-594uv^4w^3+729u^4v^4-972u^2v^6+486v^8)$$ or $f(u)\geq0,$ where $$f(u)=3u^2w^6-28v^2w^6-126u^5w^3+114u^3v^2w^3+70uv^4w^3+189u^4v^4-252u^2v^6+30v^8.$$ But $f'(u)\geq0,$ which says that it's enough to prove $f(u)\geq0$ for a minimal value of $u$, which by $uvw$ happens in equality case of two variables.
Let $b=c=1$.
Thus, we need to prove that: $$(a-1)^2(7a^4-4a^3+12a^2-8a+5)\geq0,$$ which is obvious.