Inequalities # 3

Algebra Level 3

Let $a , b , c$ be positive real numbers such that $abc = 1$ , then what is the minimum value of the expression below

$\large \sum_{\mathrm{cyc}}^{a,b,c} \dfrac{1}{a^3(b+c)}$

If your answer is of the form $\dfrac{A}{B}$ , where $A$ and $B$ are coprime positive integers , then enter the value of $A+B$ .

Source : RMO training camp(2016)

The answer is 5.

1 solution

Chew-Seong Cheong
Nov 8, 2017

$\begin{aligned} X & = \sum_{\text{cyc}}^{a,b,c} \frac 1{a^3(b+c)} & \small \color{#3D99F6} \text{Multiply up and down by }b^2c^2 \\ & = \sum_{\text{cyc}}^{a,b,c} \frac {b^2c^2}{a^3b^3c^2+a^3b^2c^3} & \small \color{#3D99F6} \text{Note that }abc = 1 \\ & = \sum_{\text{cyc}}^{a,b,c} \frac {b^2c^2}{ab+ca} & \small \color{#3D99F6} \text{By Titu's lemma} \\ & \ge \frac {(ab+bc+ca)^2}{2(ab+bc+ca)} \\ & = \frac 12(ab+bc+ca) & \small \color{#3D99F6} \text{Divided by }abc = 1 \\ & = \frac 12 \left(\frac 1a+\frac 1b+\frac 1c\right) & \small \color{#3D99F6} \text{By AM-GM inequality} \\ & \ge \frac 3{2\sqrt[3]{abc}} = \frac 32 & \small \color{#3D99F6} \text{Equality occurs when }a=b=c=1 \end{aligned}$

$\implies A+B = 3+2 = \boxed{5}$

References:

Titu's lemma
AM-GM inequality

nicely done sir!(+1)

Rishu Jaar - 3 years, 7 months ago

Inequalities # 3

The answer is 5.

1 solution

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