Two versions of AM-GM inequality

Question

I'm currently studying functional analysis and in the preliminary chapter the author gives the following inequality:

AM-GM Inequality:

Let $x,y>0$ and $0< \lambda <1$. Then $$x^\lambda y^{1-\lambda} \leq \lambda x+ (1-\lambda)y$$

But I know the AM-GM Inequality of two positive numbers $x$ and $y$ is $$\sqrt{xy} \leq \frac{x+y}{2}$$

But the first version is different. What is the difference between these two versions? Actually the latter one is the special case of the first one, namely, putting $\lambda =1/2$. Its ok.

So, why the author gives that name for the first one? Is there anything special about the first one?

Any help? and thanks in advance

The first one is weighted AM-GM inequality. – Ѕᴀᴀᴅ Oct 03 '18 at 13:05 — Ѕᴀᴀᴅ, Oct 03 '18 at 13:05
Maybe you mean $x^\lambda y^{1-\lambda}$ in the LHS. – Hugo Oct 03 '18 at 13:07 — Hugo, Oct 03 '18 at 13:07

TheSilverDoe · Accepted Answer · 2018-10-03T13:23:51.377

0

Taking the logarithm, your inequality is equivalent to $$\lambda \ln(x) + (1-\lambda) \ln(y) \leq \ln (\lambda x + (1-\lambda)y)$$

which expresses the fact that $\ln$ is concave.

The classical AM-GM inequality expresses this concavity in the special case where you are considering the middle point between $x$ and $y$. But of course the concavity is true for every convex combination of $x$ and $y$.

edited Oct 03 '18 at 13:23

answered Oct 03 '18 at 13:08

TheSilverDoe

29,720

Is there any other intuition behind this? – Oct 03 '18 at 13:18
By "exprimes," do you mean "expresses"? – Connor Harris Oct 03 '18 at 13:23
Sorry for my bad English !... I edited. – TheSilverDoe Oct 03 '18 at 13:24

Two versions of AM-GM inequality

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