An algebra problem by Vishnu Kadiri

Algebra Level 3

Let $a$ , $b$ , $c$ , and $d$ be positive reals such that $a+b+c+d=1$ . Find the maximum value of $\large \frac { ab }{ a+b } +\frac { ac }{ a+c } +\frac { ad }{ a+d } +\frac { bc }{ b+c } +\frac { bd }{ b+d } +\frac { cd }{ c+d }$

The answer is 0.75.

2 solutions

Vishnu Kadiri
Sep 24, 2017

Chew-Seong Cheong
Sep 25, 2017

Relevant wiki: Power Mean Inequality (QAGH)

$\begin{aligned} X & = \sum_{cyc} \frac {ab}{a+b} & \small \color{#3D99F6} \text{Divide up and down by }ab \\ & = \sum_{cyc} \frac {1}{\frac 1a+\frac 1b} & \small \color{#3D99F6} \text{By AM-HM inequality: }\frac {a+b}2 \ge \frac 2{\frac 1a + \frac 1b} \\ & \le \sum_{cyc} \frac {a+b}4 \end{aligned}$

$\implies X \le \dfrac {3(a+b+c+d)}4 = \dfrac 34 = \boxed{0.75}$

In the last and penultimate step, there should be an equal sign instead of smaller than or equal sign because a+b+c+d=1 not a+b+c+d>=1.

Vishnu Kadiri - 3 years, 8 months ago

I actually mean $X \le \dfrac {3(a+b+c+d}4$ and $X \le \dfrac 34$ so that it doesn't read $X = \dfrac 34$ .

Chew-Seong Cheong - 3 years, 8 months ago

Correct but it may also be read like this: 3(a+b+c+d)/4>=3/4

Vishnu Kadiri - 3 years, 8 months ago

An algebra problem by Vishnu Kadiri

The answer is 0.75.

2 solutions

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