Prove $(a_1+b_1)^{1/n}\cdots(a_n+b_n)^{1/n}\ge \left(a_1\cdots a_n\right)^{1/n}+\left(b_1\cdots b_n\right)^{1/n}$

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

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

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

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".

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!

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?

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