# Let a,b,c be positive reals such that a+b+c=1. Prove that (bc)/(a+5)+(ca)/(b+5)+(ab)/(c+5)<=1/4. Progress: This is equivalent to show that (1)/(a^2+5a)+(1)/(b^2+5b)+(1)/(c^2+5c)<=(1)/(4abc). I'm not sure how to proceed further. Also equality doesn't occur when a=b=c, which makes me more confused.

Inequality with a+b+c=1
Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that
$\frac{bc}{a+5}+\frac{ca}{b+5}+\frac{ab}{c+5}\le \frac{1}{4}.$
Progress: This is equivalent to show that
$\frac{1}{{a}^{2}+5a}+\frac{1}{{b}^{2}+5b}+\frac{1}{{c}^{2}+5c}\le \frac{1}{4abc}.$
I'm not sure how to proceed further. Also equality doesn't occur when $a=b=c$, which makes me more confused.
Edit: The original inequality is as follows: Let ${a}_{1},{a}_{2},\cdots ,{a}_{n}$ be $n>2$ positive reals such that ${a}_{1}+{a}_{2}+\cdots +{a}_{n}=1$. Prove that,
$\sum _{k=1}^{n}\frac{\prod _{j\ne k}{a}_{j}}{{a}_{k}+n+2}\le \frac{1}{\left(n-1{\right)}^{2}}.$
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Miya Swanson
Notice that the conditions imply that $0\le a,b,c\le 1$; in particular, we have
$\frac{bc}{a+5}+\frac{ca}{b+5}+\frac{ab}{c+5}\le \frac{bc}{5}+\frac{ca}{5}+\frac{ab}{5}\le \frac{b+c+a}{5}<\frac{1}{4}.$
I don't think this works for the general case though, because for $n\ge 4$, we have $\left(n-1{\right)}^{2}\ge n+2$
Now for the general case, by AM-GM inequality
$\prod _{j\ne k}{a}_{j}\le {\left(\frac{\sum _{j\ne k}{a}_{j}}{n-1}\right)}^{n-1}={\left(\frac{1-{a}_{k}}{n-1}\right)}^{n-1}\le \frac{1}{\left(n-1{\right)}^{n-1}},$
and thus
$\sum _{k=1}^{n}\frac{\prod _{j\ne k}{a}_{j}}{{a}_{k}+n+2}\le \frac{n}{\left(n+2\right)\left(n-1{\right)}^{n-1}}\le \frac{1}{\left(n-1{\right)}^{n-1}}\le \frac{1}{\left(n-1{\right)}^{2}}$
since $n-1\ge 2$
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ct1a2n4k
We'll prove a stronger inequality: $\sum _{cyc}\frac{bc}{5+a}\le \frac{1}{16}$
Indeed, we need to prove that
$\sum _{cyc}\frac{bc}{6a+5b+5c}\le \frac{a+b+c}{16}$
or
$\sum _{cyc}\left(\frac{a}{16}-\frac{bc}{6a+5b+5c}\right)\ge 0$
or
$\sum _{cyc}\frac{6{a}^{2}+5ab+5ac-16bc}{6a+5b+5c}\ge 0$
or
$\sum _{cyc}\frac{\left(a-b\right)\left(3a+8c\right)-\left(c-a\right)\left(3a+8b\right)}{6a+5b+5c}\ge 0$
or
$\sum _{cyc}\left(a-b\right)\left(\frac{3a+8c}{6a+5b+5c}-\frac{3b+8c}{6b+5a+5c}\right)\ge 0$
or
$\sum _{cyc}\frac{\left(a-b{\right)}^{2}\left(15a+15b+7c\right)}{\left(6a+5b+5c\right)\left(6b+5a+5c\right)}\ge 0.$
Done!