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I have created a new one this is the following :

Let $a,b,c>0$ such that $a^a+b^b+c^c=3$ then we have : $$a+b+c\leq 3$$

I have tried to use Jensen's inequality like this :

$$(a+b+c)\ln(\frac{a+b+c}{3})\leq a\ln(a)+b\ln(b)+c\ln(c)\leq 3\ln(\frac{a^a+b^b+c^c}{3})$$

My question is how to prove it without using Jensen's inequality ?

Thanks a lot !

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  • $\begingroup$ Why not use Jensen's inequality? $\endgroup$
    – vonbrand
    Aug 30, 2019 at 19:44

1 Answer 1

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$x \le x^x $ holds for all positive real numbers: $$ x \le x^x \iff \log x \le x \log x \iff 0 \le (x-1) \log x $$ which is certainly true. Therefore $$ a + b + c \le a^a+b^b+c^c = 3 $$ with equality exactly for $a=b=c=1$.

Your proof does not work because $$ (a+b+c)\ln(\frac{a^2+b^2+c^2}{a+b+c})\leq a\ln(a)+b\ln(b)+c\ln(c) $$ does not hold in general, try e.g. $(a, b, c) = (1, 1, 0.5)$.

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  • $\begingroup$ That's a slick one! $\endgroup$
    – AlvinL
    Aug 29, 2019 at 16:12

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