Inequality Problem

Algebra Level 4

$\dfrac{ a^3}{(1+b)(1-c)} + \dfrac{b^3}{(1+c)(1-d)} + \dfrac{c^3}{(1+d)(1-a)} +\dfrac{d^3}{(1+a)(1-b)}$

Find minimum value of above expression if $a,b,c$ and $d$ are positive real numbers such that $a+b+c+d=1$ .

The minimum value can be expressed as $\dfrac{ x}{y}$ , where $x$ and $y$ are relatively prime positive integers. Enter your answer as $x + y$ .

The answer is 16.

1 solution

Chew-Seong Cheong
Aug 11, 2017

By Hölder's inequality . we have:

$\begin{aligned} \left(\sum_{cyc} \frac {a^3}{(1+b)(1-c)} \right)^\frac 13 \left(\sum_{cyc} (1+b) \right)^\frac 13 \left(\sum_{cyc} (1-c)\right)^\frac 13 & \ge a + b + c \\ \left(\sum_{cyc} \frac {a^3}{(1+b)(1-c)} \right)^\frac 13 \left(5 \right)^\frac 13 \left(3\right)^\frac 13 & \ge 1 \\ \implies \sum_{cyc} \frac {a^3}{(1+b)(1-c)} & \ge \frac 1{15} \end{aligned}$

Equality occurs when $a=b=c=d=\frac 14$ .

$\implies x+y = 1 + 15 = \boxed{16}$

You have only shown that 1/15 is a lower bound, but is 1/15 achievable?

Pi Han Goh - 3 years, 10 months ago

Thanks, I forgot.

Chew-Seong Cheong - 3 years, 10 months ago

Inequality Problem

The answer is 16.

1 solution

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