14
$\begingroup$

consider positive numbers $a_1,a_2,a_3,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$. does the following in-equality holds and if it does then how to prove it

$\left[(a_1+b_1)(a_2+b_2)\cdots(a_n+b_n)\right]^{1/n}\ge \left(a_1a_2\cdots a_n\right)^{1/n}+\left(b_1b_2\cdots b_n\right)^{1/n}$

$\endgroup$
4
  • 1
    $\begingroup$ It does not hold when $n=1$, $a_1=b_1=3$. $\endgroup$ Mar 27, 2011 at 20:57
  • $\begingroup$ is there a typo? as it is, it is not true for large numbers $\endgroup$
    – user8268
    Mar 27, 2011 at 20:57
  • 2
    $\begingroup$ @SORRY. i missed the "+" sign on the rhs $\endgroup$
    – Mia
    Mar 27, 2011 at 21:27
  • 1
    $\begingroup$ FYI, this is known as Mahler's Inequality $\endgroup$ Jun 24, 2012 at 9:02

3 Answers 3

16
$\begingroup$

Apply the AM-GM inequality to the sequence $a_k/(a_k+b_k)$ and then to the sequence $b_k/(a_k+b_k)$. Add the resulting two inequalities, and multiply through by $\left(\prod_k (a_k+b_k)\right)^{1/n}$ to get the result.

This is exercise 2.11 (page 34) of J. Michael Steele's The Cauchy-Schwarz Master Class, and the result is there credited to Minkowski. This inequality is sometimes called the "superadditivity of the geometric mean".

$\endgroup$
3
$\begingroup$

By Holder $$\prod_{i=1}^n(a_i+b_i)\geq\left(\sqrt[n]{\prod_{i=1}^na_i}+\sqrt[n]{\prod_{i=1}^nb_i}\right)^n$$ and we are done!

$\endgroup$
2
$\begingroup$

The inequality does not hold.

Set $n=1$: the inequality reads $$a + b \geq a b.$$ With $a=2$ and $b=4$ the inequality is violated.

Why did you think the inequality does hold?

$\endgroup$
2
  • $\begingroup$ for n = 1 it reads $a+b \geq a+b$, or was the question edited? $\endgroup$
    – MichalisN
    Mar 27, 2011 at 23:42
  • $\begingroup$ It was edited an hour after Fabian answered $\endgroup$
    – Henry
    Mar 28, 2011 at 0:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.