0

Let $a, b, c$ be positive real numbers such that $abc=1$. Prove that $$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} \geqslant \frac{3}{2}.$$

Using the substitutions $\alpha=\frac{1}{a}, \beta=\frac{1}{b}, \gamma=\frac{1}{c}$ results in $\alpha\beta\gamma =1$.

Substituting to the left-hand side of the given inequality results in

$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}= \frac{\alpha^2}{\beta+\gamma}+\frac{\beta^2}{\gamma+\alpha} + \frac{\gamma^2}{\alpha+\beta}$$

so now the objective would be to prove

$$\frac{\alpha^2}{\beta+\gamma}+\frac{\beta^2}{\gamma+\alpha} + \frac{\gamma^2}{\alpha+\beta} \geqslant \frac{3}{2}.$$

How should one approach this? It looks like one could use $C-S$, but I'm not quite sure how to construct that. Also could Jensen be applied here?

3 Answers3

2

Here's a neat lemma popularly known as Titu's Lemma.

For $a,b>0$, we have the inequality $$\frac{x^2}{a}+\frac{y^2}{b}\geq \frac{(x+y)^2}{a+b}$$ The proof is easy.

One sees that $$x^2\left (1 +\frac{a}{b}\right )+y^2\left (1 +\frac{b}{a}\right )\geq x^2 +2xy +y^2 $$ by AM-GM and then you just divide both sides by $a+b$

Using this lemma twice on the LHS of your last inequality we get $$LHS\geq \frac{1}{2}(\alpha + \beta + \gamma )\geq \frac{3}{2}$$ by another application of AM-GM since $\alpha\beta\gamma=1$

user6
  • 4,092
  • Using Titu’s lemma I managed to get $\frac{\alpha^2}{\beta + \gamma} + \frac{\beta^2}{\gamma + \alpha} + \frac{\gamma^2}{\alpha + \beta} \geqslant \frac{(\alpha +\beta + \gamma )^2}{2(\alpha + \beta + \gamma)} $ and from $AM-GM$ I got an bound for $\alpha + \beta + \gamma \geqslant 3.$ Shouldn’t this work as well? –  May 17 '20 at 21:57
  • 1
    Yeah that's correct. – user6 May 17 '20 at 22:10
1

Well, if you wanted to use C-S, here's how

$\alpha + \beta+ \gamma \ge3\sqrt[3]{\alpha\beta\gamma}=3$, so $$\left(\frac{\alpha^2}{\beta+ \gamma}+\frac{\beta^2}{\alpha +\gamma}+\frac{\gamma^2}{\alpha +\beta}\right) ((\beta+\gamma)+(\alpha +\gamma)+(\alpha +\beta)) \ge(\alpha +\beta+\gamma)^2$$ $$\iff \left(\frac{\alpha^2}{\beta+ \gamma}+\frac{\beta^2}{\alpha +\gamma}+\frac{\gamma^2}{\alpha +\beta}\right) (2(\alpha+\beta+\gamma)) \ge(\alpha +\beta+\gamma)^2$$ $$\iff \frac{\alpha^2}{\beta+ \gamma}+\frac{\beta^2}{\alpha +\gamma}+\frac{\gamma^2}{\alpha +\beta} \ge\frac{\alpha +\beta+\gamma}{2} \ge \frac{3}{2}$$

Anas A. Ibrahim
  • 1,884
  • 7
  • 15
0

Also, by Holder and AM-GM we obtain: $$\sum_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{b^3c^3}{b+c}\geq\frac{(ab+ac+bc)^3}{3\sum\limits_{cyc}(a+b)}\geq\frac{3abc(a+b+c)\cdot3\sqrt{a^2b^2c^2}}{6(a+b+c)}=\frac{3}{2}.$$

Michael Rozenberg
  • 194,933