# Prove that $\dfrac{a}{b^2+5}+ \dfrac{b}{c^2+5} + \dfrac{c}{a^2+5} \le \dfrac 12$

Let $a,b,c>0$ and $a^3+b^3+c^3=3$. Prove that $$\dfrac{a}{b^2+5}+ \dfrac{b}{c^2+5} + \dfrac{c}{a^2+5} \le \dfrac 12$$ I have an ugly solution for this solution.

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have you tried Cauchy-Schwarz inequality? –  user42912 Feb 17 at 7:47
No. I don't know how to try Cauchy-Schwarz inequality. Can you post your solution in here ? Thank you. –  boxmath Feb 17 at 7:50
I don't have time to do this now, please see this book and search for Cauchy-schwarz reverse technique. google.com/… –  user42912 Feb 17 at 8:20
@user42912: Thank you very much for the book! –  Salech Alhasov Feb 17 at 8:35
@SalechAlhasov you're welcome, I'm glad to help! –  user42912 Feb 17 at 8:39

First ,we can get rid of denominators, by noticing that $\frac{1}{x^2+5} \leq \frac{4-x}{18}$ for any $x\in [0,2]$ (indeed, we have $$(4-x)(x^2+5)-18=((x-1)^2)(2-x) \geq 0 \tag{1}$$

So it will suffice to show that the number

$$f(a,b,c)=a(4-b)+b(4-c)+c(4-a) \tag{2}$$

is $\leq 9$. Note that $f(a,b,c)=4(a+b+c)-(ab+ac+bc)$ is fully symmetric in $a,b,c$.

We need a “lower-dimensional” result :

Lemma. Let $\alpha,\beta$ be two positive numbers such that $\alpha^3+\beta^3=2$. Then $\alpha(3-\beta)+\beta(3-\alpha) \leq 4$, with equality iff $\alpha=\beta=1$.

Proof of lemma The proposed inequality is equivalent to $\beta(3-2\alpha) \leq 4-3\alpha$, or $(2-\alpha^3)(3-2\alpha)^3 \leq (4-3\alpha)^3$. Now

$$(4-3\alpha)^3-(2-\alpha^3)(3-2\alpha)^3=(\alpha-1)^4 \bigg(\frac{42}{25}+ \big(\frac{13}{10}-\alpha\big)\big(8\alpha+\frac{32}{5}\big)\bigg),$$

which concludes the proof of the lemma.

Now let $a,b,c$ be positive numbers with $a^3+b^3+c^3=1$. Set $t=\big(\frac{b^3+c^3}{2}\big)^{\frac{1}{3}}$, so that $b^3+c^3=2t^3$. Then the numbers $\alpha=\frac{b}{t}$ and $\beta=\frac{c}{t}$ satisfy the hypotheses of the lemma ; we deduce

$$b(3t-c)+c(3t-b) \leq 4t^2, \ \text{with equality iff } \ b=c=t \tag{3}$$

This means that

$$f(a,b,c) \leq f(a,t,t), \ \text{with equality iff } \ b=c=t \tag{4}$$

The set $K=\lbrace (a,b,c) \in [0,+\infty[^3 | a^3+b^3+c^3=1\rbrace$ is compact, so the continuous map $f$ attains its maximum on $K$ at some point $(a_0,b_0,c_0)$. Then (4) shows that we must have $b_0=c_0$, and by symmetry $a_0=b_0=c_0=1$, qed.

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