2
$\begingroup$

My question may looks very simple:

I know that $(a+b)^{2}\leq 2(a^{2}+b^{2})$ for any $a,b\in \mathbb R$. Do we have an inequality like $(a+b)^{2} \leq C\, a^{p}b^{p}$ for some constant $C$ depends only on the powers $2,p>0$?

Edit: $0 .

  • 0
    @Nichole Take a look at [Jensen's inequality](http://en.wikipedia.org/wiki/Jensen%27s_inequality). The logarithm function is concave, so it happens to be exactly the wrong way.2012-06-24

1 Answers 1

1

Attempting to answer the question in the last comment.

For positive real numbers $a,b$ we have the AM-GM inequality $ \frac{a+b}2\ge\sqrt{ab}. $ In the case of two numbers this is easy to prove by squaring both sides and juggling the terms a bit. Taking logarithms of this gives after straightforward manipulations (assuming that the base of the logarithm is $>1$ so that the logarithm is an increasing function) $ \log(a+b)\ge \frac{\log a+\log b-\log 4}2. $ It is easy to see that we have equality here if and only if $a=b$.

As you seem to want an upper bound to this may mean a "back to the drawing board"-moment for you, but them's the breaks.