Classical

Algebra Level 3

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

Let $a$ , $b$ , and $c$ be positive real numbers such that $abc=1$ , then find the minimum value of the expression above.

1 solution

Aditya Narayan Sharma
Jun 9, 2016

Put $\displaystyle a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$ and it turns,

$\displaystyle S = \sum_{x,y,z} \frac{x^3 yz}{y+z} = xyz\sum_{x,y,z} \frac{x^2}{y+z}$

Since we have $xyz=1$ , $\displaystyle x+y+z \ge 3$

$\displaystyle \sum_{x,y,z} \frac{x^2}{y+z} \ge \frac{(x+y+z)^2}{2(x+y+z)} \ge \frac{3}{2}$

So , $\displaystyle S=\sum_{a,b,c}\frac{1}{a^3(b+c)} = \sum_{x,y,z} \frac{x^3 yz}{y+z} \ge \frac{3}{2}$

Interesting Solution! +1

Mehul Arora - 5 years ago

Aditya Narayan Sharma - 5 years ago