using a=1-b-c b=1-c-a c=1-a-b we can getΣ1/(a+bc)=2/(a+b)(c+a)(b+c) Then we only need to prove 2>=7Σa^2b+7Σab^2+12abc using2=2(a+b+c)^3=2Σa^3+6Σa^2b+6Σab^2+12abc we only need to prove 2Σa^3>=Σa^2b+Σab^2 which is quite obvious

@above I wrote your sol in LaTeX. Using $a=1-b-c, \ b=1-c-a$ and $c=1-a-b$ we can get \[\sum\frac{1}{a+bc}=\frac{2}{(a+b)(c+a)(b+c)}.\]Therefore, we only need to prove \[\frac{2}{(a+b)(c+a)(b+c)}\geq \frac{7}{1+abc}\iff 2\geq 7\sum ab(a+b)+12abc.\]However, because $$2=2(a+b+c)^3=2\sum a^3+6\sum ab(a+b)+12abc$$we only need to prove that $$2\sum a^3\geq \sum ab(a+b)$$which is quite obvious.

Here's a simpler solution just by manipulation and CS This is equivalent to proving \[ \sum \frac{1 + abc}{a + bc} \ge 7 \Longleftrightarrow \sum \frac{(a + 1)(bc + 1)}{(a + b)(a + c)} \ge 10 \]However, \begin{align*} \sum \frac{(a + 1)(bc + 1)}{(a + b)(a + c)} &= \sum \frac{(a + b + a + c)(bc + 1)}{(a + b)(a + c)} \\ &= \sum \frac{bc + 1}{a + c} + \frac{bc + 1}{a + b} \\ &= \sum \frac{bc + 1}{a + c} + \frac{ba + 1}{a + c} \\ &= \sum b + 2 \sum \frac{1}{a + c} \\ &= 1 + 2 \sum \frac{1}{1 - b} \\ &\ge 1 + 2 \cdot \frac{9}{3 - (a + b + c)} \\ &= 10 \end{align*}

Let $a,b,c> 0$ and $a+b+c=3.$ Prove that$$\frac{3}{2}\leq\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}\leq \frac{3}{abc+1}$$Let $a,b,c>0 , a+b+c=1$ . Prove that$$\frac{27}{4}\leq\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}\leq \frac{2}{\sqrt{3abc}-abc}$$$$\frac{a}{b+ca}+\frac{b}{c+ab}+\frac{c}{a+bc}\leq\frac{7}{3(abc+1)}$$Let $a,b,c\geq0,a+b+c=1$. Prove that \[\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}\geq\frac{8}{1+5abc}.\]h h h

oVlad wrote: Let $a,b,c>0$ be real numbers with the property that $a+b+c=1$. Prove that \[\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}\geq\frac{7}{1+abc}.\] Let $a= \frac{x}{x+y+z},b=\frac{y}{x+y+z}, c= \frac{z}{x+y+z}$ where $x,y$ and $z$ are positive real numbers. Then, $\sum_{cyc} \frac{1}{a+bc} = (x+y+z)^2 \sum_{cyc} \frac{1}{(x+y)(x+z)}= \frac{2(x+y+z)^3}{(x+y)(x+z)(y+z)}$ and $\frac{7}{1+abc}= \frac{7(x+y+z)^3}{(x+y+z)^3+xyz}$ So it's equivalent to prove $\frac{2}{(x+y)(y+z)(z+x)} \geq \frac{7}{(x+y+z)^3+xyz} \iff 2(x+y+z)^3+2xyz \geq 7(x+y+z)(xy+xz+yz)-7xyz.$ But it's true since $(x+y+z)^3+9xyz \geq 4(x+y+z)(xy+xz+yz)$ by Schur's and $(x+y+z)^3 \geq 3(x+y+z)(xy+xz+yz)$ by AM-GM.

oVlad wrote: Let $a,b,c>0$ be real numbers with the property that $a+b+c=1$. Prove that \[\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}\geq\frac{7}{1+abc}.\] Solution of Zhangyanzong: \[\sum\frac{1}{a+bc}=\frac{2}{(a+b)(c+a)(b+c)}=\frac{2}{\sum a \sum ab-abc} \]\[\sum\frac{1}{a+bc}\geq\frac{7}{1+abc}\iff 2(1+abc)\geq 7(\sum a \sum ab-abc)\iff 2+9abc\geq 7\sum ab\]$$(\sum a )^3+abc\geq4 \sum a \sum ab \implies$$$$ 2+9abc\geq 1+4\sum ab=(\sum a )^2+4\sum ab\geq 7\sum ab$$ Let $a,b,c>0$ and $a+b+c+2\leq abc$ . Prove that $$\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}\leq \frac{1}{2}.$$ sqing wrote: Let $a,b,c> 0$ and $a+b+c=1.$ Prove that $$\frac{a}{b+ca}+\frac{b}{c+ab}+\frac{c}{a+bc}\leq\frac{7}{3(abc+1)}$$

Here is a "solution" by UVW:

By T2 $LHS\ge \frac{9}{1+\sum ab} \ge RHS$ IF AND ONLY IF $2+9abc\ge 7\sum ab$ Which is just JBMO 2024 SL A2

My solution is almost the exact same to the above solution, but I'm just posting this for storage Note that $\text{LHS} = \sum_{cyc} \frac{1}{a(a+b+c)+bc} = \sum_{cyc} \frac{1}{(a+b)(a+c)} = \frac{2}{(a+b)(a+c)(b+c)}.$ It thus suffices to show that $$\frac{2}{(a+b)(a+c)(b+c)} \geq \frac{7}{1+abc}.$$Eliminating fractions and then homogenizing, we find that Muirhead finishes.