Prove an inequality without using Jensen's inequality

Question

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 !

Answer 1

$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 $a=b=c=1$.

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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. $(a, b, c) = (1, 1, 0.5)$.