5

For three positive real numbers $a,b,$ and $c$, prove that $$\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b} \geq 1.$$

Attempt

Rewritting we obtain $\dfrac{2 a^3+2 a^2 b-3 a^2 c-3 a b^2-3 a b c+2 a c^2+2 b^3+2 b^2 c-3 b c^2+2 c^3}{(a+2b)(2a+c)(b+2c)} \geq 0$. Then to I proveed to use rearrangement, AM-GM, etc. on the numerator?

Martin Sleziak
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Jacob Willis
  • 1,601

2 Answers2

7

Very similar to Nesbitt's inequality. If we set $A=b+2c,B=c+2a,C=a+2b$, we have $ 4A+B-2C = 9c $ and so on, and the original inequality can be written as: $$ \frac{4B+C-2A}{9A}+\frac{4C+A-2B}{9B}+\frac{4A+B-2C}{9C} \geq 1 $$ or: $$ \frac{4B+C}{A}+\frac{4C+A}{B}+\frac{4A+B}{C} \geq 15 $$ that follows from combining $\frac{B}{A}+\frac{A}{B}\geq 2$ (consequence of the AM-GM inequality) with $\frac{B}{A}+\frac{C}{B}+\frac{A}{C}\geq 3$ (consequence of the AM-GM inequality again).

Jack D'Aurizio
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4

Use Cauchy–Schwarz's inequality as follows :

$$(a(b+2c)+b(c+2a)+c(a+2b)) \left (\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \right ) \geq (a+b+c)^2$$ and then :

$$\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \geq \frac{(a+b+c)^2}{3(ab+bc+ca)}$$

Now it's easy to see that $$\frac{(a+b+c)^2}{3(ab+bc+ca)} \geq 1$$ this being equivalent with $$a^2+b^2+c^2 \geq ab+bc+ca$$ which is in turn equivalent with $$(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$$ (after a multiplication with $2$ and rearranging )