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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

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    Well, it might be the case that $a = 0, b \neq 0$, then left side is positive, but right is not.2012-06-24
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    This is impossible due to the case $a=0$ (or $a\to0$).2012-06-24
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    Ok, so I should add that $0< a,b$, thank you.2012-06-24
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    @Nichole that won't help, you can still let $a\rightarrow 0$ and the rhs will be arbitrarily close to $0$.2012-06-24
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    Ok, may be this will help: I'm trying to find some relation between $\log(a+b)$ and $\log a +\log b$ for $a,b>0$ since we don't have inequality like $\log(a+b)\leq \log a+\log b$.2012-06-24
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    @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

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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.