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Let $a$ and $b$ be two positive numbers such that $a+b=1$. I am supposed to show that $u^av^b\leq au+bv$ for all positive $u$ and $v$.

It is known that $\ln x \leq x$ for all positive $x$, so I managed to get $\ln(u^av^b)=a\ln(u)+b\ln(v)\leq au+bv$ instead. Not sure if this direction is worth exploring or did I miss the point altogether?

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  • $\begingroup$ But I got stuck since $\ln(u^av^b) \leq u^av^b$ instead of $\ln(u^av^b) \geq u^av^b$. Would appreciate some pointers if possible. $\endgroup$
    – KHOOS
    Dec 2, 2022 at 0:27
  • $\begingroup$ Hint: use Jensen's inequality to bound $a \log u + b\log v$. $\endgroup$
    – Damian Pavlyshyn
    Dec 2, 2022 at 0:49
  • $\begingroup$ Thank you! I got it! That is a very helpful hint. Have a good day. $\endgroup$
    – KHOOS
    Dec 2, 2022 at 1:00
  • $\begingroup$ Identical duplicate in less than 1 hour: Sequence of geometric mean subtracted by arithmetic mean $\endgroup$
    – Ѕᴀᴀᴅ
    Dec 2, 2022 at 2:12

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