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Q: For positive reals $a$, $b$ and $c$, prove that $a^3b + b^3c + c^3a \ge abc(a+b+c)$.

Ash_Blanc For positive reals $a$, $b$ and $c$, prove that $a^3b + b^3c + c^3a \ge abc(a+b+c)$. Well Chebysev's inequality looks natural here though it doesn't seems that easy to find the right order of $(a^2b, b^2c, c^2a)$ like if W.L.O.G we assume $a \ge b \ge c$ then clearly $a^2b \ge b^2c$ but how to ...

Nov 1, 2015 at 21:02, by Martin Sleziak
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Nov 1, 2015 at 21:02, by Martin Sleziak
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Q: How are the following two Chebychev's inequalities equivalent?

nbroI was looking at the following definition of Chebyshev's inequality $$P(|X - E(X)| \geq r) \leq \frac{Var(X)}{r^2}$$ which includes the expected value and variance of $X$, and then I discovered there's another equivalent Chebyshev's inequality, which involves the standard deviation $\sigma$ $$...

Nov 1, 2015 at 21:02, by Martin Sleziak
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Q: Lower bound inequality for $P(X<\mu-\sigma)$

stackExchangeUserIs there a lower bound inequality for $P(X<\mu-\sigma)$, where $P$, $\mu$ and $\sigma$ might be arbitrary but finite. For example by the Cantelli's inequality we have an upper bound $P(X<\mu-\sigma)\le0.5$. Is there also a nontrivial lower bound inequality?


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