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Prove that $$a^ab^bc^c\geq \left(\frac{a+b}{2}\right)^{\frac{a+b}{2}} \left(\frac{b+c}{2}\right)^{\frac{b+c}{2}}\left(\frac{c+a}{2}\right)^{\frac{c+a}{2}}\geq \left(\frac{a+b+c}{3}\right)^{a+b+c}$$ where $a,b,c$ are positive real numbers.

I am able to prove the $\left(\frac{a+b}{2}\right)^{\frac{a+b}{2}} \left(\frac{b+c}{2}\right)^{\frac{b+c}{2}}\left(\frac{c+a}{2}\right)^{\frac{c+a}{2}}\geq \left(\frac{a+b+c}{3}\right)^{a+b+c}$ by using weighted AM$\geq$ GM on the numbers $\frac{a+b}{2}, \frac{b+c}{2}, \frac{c+a}{2}$ with associated weights $\frac{a+b}{2}, \frac{b+c}{2}, \frac{c+a}{2}$.

The problem is to prove $$\bbox[5px,border:2px solid #C0A000]{a^ab^bc^c\geq \left(\frac{a+b}{2}\right)^{\frac{a+b}{2}} \left(\frac{b+c}{2}\right)^{\frac{b+c}{2}}\left(\frac{c+a}{2}\right)^{\frac{c+a}{2}}}$$ Please help me to solve it.

user1942348
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1 Answers1

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For $x > 0 \Rightarrow f(x) = x\ln x \Rightarrow f'(x) = 1 + \ln x \Rightarrow f''(x) = \dfrac{1}{x} > 0\Rightarrow f(a)+ f(b) \geq 2f\left(\dfrac{a+b}{2}\right)\Rightarrow a\ln a + b\ln b \geq 2\left(\dfrac{a+b}{2}\right)\ln\left(\dfrac{a+b}{2}\right)\Rightarrow \ln(a^ab^b) \geq \ln\left(\left(\dfrac{a+b}{2}\right)^{a+b}\right)\Rightarrow a^ab^b \geq \left(\dfrac{a+b}{2}\right)^{a+b}$. Repeat this process for the other pairs $(b,c)$,and $(c,a)$ we have: $a^ab^bb^bc^cc^ca^a \geq \left(\dfrac{a+b}{2}\right)^{a+b}\left(\dfrac{b+c}{2}\right)^{b+c}\left(\dfrac{c+a}{2}\right)^{c+a}\Rightarrow (a^ab^bc^c)^2 \geq \left(\left(\dfrac{a+b}{2}\right)^{\frac{a+b}{2}}\left(\dfrac{b+c}{2}\right)^{\frac{b+c}{2}}\left(\dfrac{c+a}{2}\right)^{\frac{c+a}{2}}\right)^2\Rightarrow \text{QED}$

DeepSea
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