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Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that $$\dfrac{bc}{a+5}+\dfrac{ca}{b+5}+\dfrac{ab}{c+5}\le \dfrac{1}{4}.$$

Progress: This is equivalent to show that $$\dfrac{1}{a^2+5a}+\dfrac{1}{b^2+5b}+\dfrac{1}{c^2+5c}\le \dfrac{1}{4abc}.$$ 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,

$$\sum_{k=1}^n \dfrac{\prod\limits_{j\ne k}a_j}{a_k+n+2}\le \dfrac{1}{(n-1)^2}.$$

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    from where does that come? – 2017-01-08
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    It's a problem from an olympiad handout – 2017-01-08
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    can you send me a copy from this handout please? – 2017-01-08
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    and how is the original inequality? – 2017-01-08

3 Answers 3

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Notice that the conditions imply that $0\leq a,b,c\leq 1$; in particular, we have $$\dfrac{bc}{a+5}+\dfrac{ca}{b+5}+\dfrac{ab}{c+5}\le \dfrac{bc}{5}+\dfrac{ca}{5}+\dfrac{ab}{5} \leq \frac{b+c+a}{5} < \dfrac{1}{4}.$$ I don't think this works for the general case though, because for $n\geq 4$, we have $(n-1)^2\geq n+2$.

Now for the general case, by AM-GM inequality $$\prod_{j\neq k} a_j \leq \left(\frac{\sum_{j\neq k}a_j}{n-1}\right)^{n-1}=\left(\frac{1-a_k}{n-1}\right)^{n-1}\leq \frac{1}{(n-1)^{n-1}},$$ and thus $$\sum_{k=1}^n \dfrac{\prod_{j\ne k}a_j}{a_k+n+2}\leq \frac{n}{(n+2)(n-1)^{n-1}} \leq \frac{1}{(n-1)^{n-1}}\leq \dfrac{1}{(n-1)^2}$$since $n-1\geq 2$.

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We'll prove a stronger inequality: $\sum\limits_{cyc}\frac{bc}{5+a}\leq\frac{1}{16}$.

Indeed, we need to prove that $$\sum\limits_{cyc}\frac{bc}{6a+5b+5c}\leq\frac{a+b+c}{16}$$ or $$\sum\limits_{cyc}\left(\frac{a}{16}-\frac{bc}{6a+5b+5c}\right)\geq0$$ or $$\sum\limits_{cyc}\frac{6a^2+5ab+5ac-16bc}{6a+5b+5c}\geq0$$ or $$\sum\limits_{cyc}\frac{(a-b)(3a+8c)-(c-a)(3a+8b)}{6a+5b+5c}\geq0$$ or $$\sum\limits_{cyc}(a-b)\left(\frac{3a+8c}{6a+5b+5c}-\frac{3b+8c}{6b+5a+5c}\right)\geq0$$ or $$\sum\limits_{cyc}\frac{(a-b)^2(15a+15b+7c)}{(6a+5b+5c)(6b+5a+5c)}\geq0.$$ Done!

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$\max(bc)=\frac14$ when $b=c=\frac12$ Then you have $a=0$, so the first term the maximum value is $\frac1{20}$. Even if you have three terms, the value is $\frac3{20}$ which is smaller than $\frac14$