Nice problem. Note that $a+bc=a(a+b+c)+bc=(a+b)(a+c)$; and $a+1=2a+b+c=(a+b)+(a+c)$. With this, we have \[ \sum \frac{a+1}{\sqrt{a+bc}} = \sum \frac{(a+b)+(a+c)}{\sqrt{(a+b)(a+c)}}\ge 6, \]since $(a+b)+(a+c)\ge 2\sqrt{(a+b)(a+c)}$ by the AM-GM inequality. Finally, using once again $a+b+c=1$, it suffices to show \[ 6\ge \frac{2(a+b+c)^2}{a^2+b^2+c^2}\iff 3(a^2+b^2+c^2)\ge (a+b+c)^2\iff a^2+b^2+c^2\ge ab+bc+ca, \]which is obvious. The equality holds iff $a=b=c=1/3$ throughout.

Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1$. Show that $$ \frac{2}{ab+bc+ca}\geq \frac{a+1}{\sqrt{a+bc}}+\frac{b+1}{\sqrt{b+ca}}+\frac{c+1}{\sqrt{c+ab}} \geq \frac{2}{a^2+b^2+c^2}$$$$ \frac{3k+1}{2(ab+bc+ca)}\geq \frac{a+k}{\sqrt{a+bc}}+\frac{b+k}{\sqrt{b+ca}}+\frac{c+k}{\sqrt{c+ab}} \geq \frac{3k+1}{2(a^2+b^2+c^2)}$$Where $k\geq 0.$

augustin_p wrote: Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1$. Show that $\frac{a+1}{\sqrt{a+bc}}+\frac{b+1}{\sqrt{b+ca}}+\frac{c+1}{\sqrt{c+ab}} \geq \frac{2}{a^2+b^2+c^2}$. When does the equality take place? By AM-GM, it holds $$ \frac{a+1}{\sqrt{a+bc}}\geq \frac{a+1}{\sqrt{a+\left(\frac{b+c}{2}\right)^2}}=\frac{a+1}{\sqrt{a+\left(\frac{1-a}{2}\right)^2}}=\frac{2(a+1)}{\sqrt{4a+1-2a+a^2}}=2 $$Therefore, $$ \sum_{cyc}\frac{a+1}{\sqrt{a+bc}}\geq 2+2+2=6 $$Hence it suffices to show $$ 6\geq \frac{2}{a^2+b^2+c^2}, $$which is inmediate by QM-AM on $a,b,c$.

Old idea We know $bc+a(a+b+c) = (a+b)(a+c)$ ( ) Also we know $a+1 = a+a+b+c = (a+b)+(a+c)$ ( ) By ( ) and ( ) we get : $$\frac{a+1}{\sqrt{a+bc}}+\frac{b+1}{\sqrt{b+ca}}+\frac{c+1}{\sqrt{c+ab}} =\frac{a+1}{\sqrt{(a+b)(a+c)}}+\frac{b+1}{\sqrt{(a+b)(a+c)}}+\frac{c+1}{\sqrt{(a+b)(a+c)}} = \frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}$$ $\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}} \geq \sum \frac{ 2\sqrt{(a+b)(a+c)}}{\sqrt{(a+b)(a+c)}} = 6$ Now we need prove $3(a^2+b^2+c^2) \geq 1 $ and we know it is true by C-S $\blacksquare$

Positive real numbers $a$, $b$, $c$ satisfy $a+b+c=1$. Show that $$\frac{a+3}{\sqrt{a+bc}}+\frac{b+3}{\sqrt{b+ca}}+\frac{c+3}{\sqrt{c+ab}} \geq \frac{5}{a^2+b^2+c^2} $$

strong_boy wrote: Old idea We know $bc+a(a+b+c) = (a+b)(a+c)$ ( ) Also we know $a+1 = a+a+b+c = (a+b)+(a+c)$ ( ) By ( ) and ( ) we get : $$\frac{a+1}{\sqrt{a+bc}}+\frac{b+1}{\sqrt{b+ca}}+\frac{c+1}{\sqrt{c+ab}} =\frac{a+1}{\sqrt{(a+b)(a+c)}}+\frac{b+1}{\sqrt{(a+b)(a+c)}}+\frac{c+1}{\sqrt{(a+b)(a+c)}} = \frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}$$ $\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}} \geq \sum \frac{ 2\sqrt{(a+b)(a+c)}}{\sqrt{(a+b)(a+c)}} = 6$ Now we need prove $3(a^2+b^2+c^2) \geq 1 $ and we know it is true by C-S $\blacksquare$ It looks nice

mihaig wrote: strong_boy wrote: Old idea We know $bc+a(a+b+c) = (a+b)(a+c)$ ( ) Also we know $a+1 = a+a+b+c = (a+b)+(a+c)$ ( ) By ( ) and ( ) we get : $$\frac{a+1}{\sqrt{a+bc}}+\frac{b+1}{\sqrt{b+ca}}+\frac{c+1}{\sqrt{c+ab}} =\frac{a+1}{\sqrt{(a+b)(a+c)}}+\frac{b+1}{\sqrt{(a+b)(a+c)}}+\frac{c+1}{\sqrt{(a+b)(a+c)}} = \frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}$$ $\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}}+\frac{(a+b)(a+c)}{\sqrt{(a+b)(a+c)}} \geq \sum \frac{ 2\sqrt{(a+b)(a+c)}}{\sqrt{(a+b)(a+c)}} = 6$ Now we need prove $3(a^2+b^2+c^2) \geq 1 $ and we know it is true by C-S $\blacksquare$ It looks nice Thanks!