Let $a,b,c>0$ such that $a^2+b^2+c^2=3.$ Prove or disprove that
$$\dfrac{\sqrt{b}+\sqrt{c}}{\sqrt{a^2+4b^2+4c^2}}+\dfrac{\sqrt{c}+\sqrt{a}}{\sqrt{b^2+4c^2+4a^2}}+\dfrac{\sqrt{{\color{red}a}}+\sqrt{{\color{red}b}}}{\sqrt{c^2+4a^2+4b^2}}\le 2.$$
For $a=b=c=1,$ LHS $=$ RHS, so, using the Cauchy-Schwarz inequality,
$$(a^2+4b^2+4c^2)(1+4+4)\ge (a+4b+4c)^2$$
so we need show
$$\sum_{cyc}\dfrac{\sqrt{b}+\sqrt{c}}{a+4b+4c}\le\dfrac{2}{3}$$
it seems that this inequality doesn't hold.
Best Answer
Using $u + v \le \sqrt{2(u^2 + v^2)}$, we have $$\sqrt b + \sqrt c \le \sqrt{2(b + c)} \le \sqrt{2\sqrt{2(b^2 + c^2)}}.\tag{1}$$ It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{\sqrt{2\sqrt{2(b^2 + c^2)}}}{\sqrt{a^2+4b^2+4c^2}} \le 2. \tag{2}$$
Using $u + v + w \le \sqrt{3(u^2 + v^2 + w^2)}$, we have $$\sum_{\mathrm{cyc}} \frac{\sqrt{2\sqrt{2(b^2 + c^2)}}}{\sqrt{a^2+4b^2+4c^2}} \le \sqrt{3\sum_{\mathrm{cyc}} \frac{2\sqrt{2(b^2 + c^2)}}{a^2 + 4b^2 + 4c^2}}.\tag{3}$$ It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{6\sqrt{2(b^2 + c^2)}}{a^2 + 4b^2 + 4c^2} \le 4. \tag{4}$$
Let $x = a^2, y = b^2, z = c^2$. We have $x + y + z = 3$. We need to prove that $$\sum_{\mathrm{cyc}} \frac{6\sqrt{2(3-x)}}{12 - 3x} \le 4. \tag{5}$$
Let $3 - x = 2u^2, 3 - y = 2v^2, 3 - z = 2w^2$. We have $u^2 + v^2 + w^2 = 3$. We need to prove that $$\frac{u}{2u^2 + 1} + \frac{v}{2v^2 + 1} + \frac{w}{2w^2 + 1} \le 1. \tag{6}$$ This is discussed and proved in https://artofproblemsolving.com/community/c6h2195486p16486899.
Remarks: We may directly prove (5) using calculus. But my proof is not nice.