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Let $a,b,c$ be positive real numbers such that $a^a b^b c^c =1$ , then is it true that $a+b+c \le 3$ ?

Souvik Dey
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  • Use Lagrange multipliers to maximise $a+b+c$ subject to the conditions $a,b,c>0$ and $a\log a+b\log b+c\log c=0$. If I have done the calculations correctly, the maximum occurs when $a=b=c=1$. – David Jun 12 '14 at 05:06

1 Answers1

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Let $f(x)=x \log x$ , for $x>0$. Clearly $f''(x)=\frac{1}{x}>0~$, so by Jensen's inequality we have $$\frac{a+b+c}{3}\log\left(\frac{a+b+c}{3}\right)\leq\frac{1}{3}(a\log a+b\log b+c\log c)= \frac{\log(a^ab^bc^c)}{3}\leq 0$$ it follows that $\log\left(\frac{a+b+c}{3}\right)\leq0$ that is $a+b+c\leq 3$.

Omran Kouba
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