For positive $a$, $b$, $c$ with $a+b+c=1$, show that $(ab+bc+ca) \sum_{cyc}\frac{a}{b^2+b} \geq \frac34$

Question

If $a,b,c > 0$ and $a+b+c = 1$, then prove that $$\left(\frac{a}{b^2+b}+\frac{b}{c^2+c}+\frac{c}{a^2+a}\right)(ab+bc+ca)\geq\frac{3}{4}$$

It's been more than 35 years since I last touched algebra!!

How did this problem arise? It looks like a contest problem ... — Noah Schweber, Jul 21 '19 at 14:58
@NoahSchweber no idea :) A friend gave it to me, challenging me to solve it. I don't know what to do, other than starting multiplications etc - I assume at some point I must use cauchy, but I don't even remember these things, sorry! — Tom Galle, Jul 21 '19 at 15:05
i've tried partial fractions and the $(a+b+c)^2 $ identity . All to no avail... — MaximusFastidiousIrreverence, Jul 21 '19 at 15:27

score 5 · Accepted Answer · answered Jul 21 '19 at 17:07

5

Hint: Using generalised Holder’s inequality
$$\sum_{cyc}ab \cdot \sum_{cyc} \frac{a}{b^2+b}\cdot \sum_{cyc} a(b+1) \geqslant \left(\sum_{cyc}a\right)^3=1$$

Now it is enough to show $\sum_{cyc} a(b+1) \leqslant \frac43$ which is easy.

answered Jul 21 '19 at 17:07

Macavity

46,381

Do you mean $a(b+1)$ or $\frac{b+1}{a}$? – EuxhenH Jul 21 '19 at 17:21
1

@EuxhenH The former - exactly as written. – Macavity Jul 21 '19 at 17:22

For positive $a$, $b$, $c$ with $a+b+c=1$, show that $(ab+bc+ca) \sum_{cyc}\frac{a}{b^2+b} \geq \frac34$

1 Answers1