To prove the following inequalities of positive rational numbers

Question

I have to prove the following inequalities:

$$ a^ab^bc^c \ge \ (\frac{a+b}{2})^{\frac{a+b}{2}} (\frac{c+b}{2})^{\frac{c+b}{2}} (\frac{a+c}{2})^{\frac{a+c}{2}} $$

$$(a+b)^{c}(c+b)^{a}(a+c)^{b} < \left(\frac{2(a+b+c)}{3}\right)^{a+b+c} ,a\ne b\ne c. $$

For the second I tried using weighted $AM > GM$ method on $a+b, b+c, c+a$ with weights $ c, a, b$.

But I am not getting the desired answer.

Answer 1

I have been able to prove the first inequality by using weighted $AM \ge\ GM$ on $ \frac{a+b}{a}, \frac{a+b}{b}, \frac{b+c}{b}, \frac{b+c}{c}, \frac{c+a}{c}, \frac{a+c}{a} $ with weights $ \frac{a}{2}, \frac{b}{2}, \frac{b}{2}, \frac{c}{2}, \frac{c}{2}, \frac{a}{2}$ respectively.

Answer 2

Another approach for the first question

$f(x)=x\ln x$ is convex for all positive $x$.

Note that taking $a\ge b\ge c\,$ (WLOG) gives us $\frac{a+b}{2}\ge \frac{a+c}{2} \ge \frac{b+c}{2}$ and

\begin{align} a+b+c &= \frac{a+b}{2}+\frac{a+c}{2}+\frac{b+c}{2} \\ a+b &\ge \frac{a+b}{2}+\frac{a+c}{2} \\ a &\ge \frac{a+b}{2}\\ \end{align}

Thus,

$$(\frac{a+b}{2}, \frac{a+c}{2}, \frac{b+c}{2}) \prec \left(a,b,c\right) $$

Hence by Karamata's Majorization Inequality,

\begin{align} f(a)+f(b)+f(c)&\ge f(\frac{a+b}{2})+f(\frac{a+c}{2})+f(\frac{b+c}{2})\\ \iff \;\;\; a \ln a + b\ln b + c \ln c &\ge (\dfrac{a+b}{2}) \ln \frac{a+b}{2} + (\dfrac{a+c}{2})\ln \frac{a+c}{2} + (\dfrac{b+c}{2}) \ln \frac{b+c}{2}\\ \iff \;\;\; a^ab^bc^c &\ge \ (\frac{a+b}{2})^{\frac{a+b}{2}} (\frac{a+c}{2})^{\frac{a+c}{2}} (\frac{b+c}{2})^{\frac{b+c}{2} } \end{align}

Hint for the second question

Let $k=a+b+c\,$. $\;f(x)=(k-x) \ln x\;$ is strictly concave for all $x \in [0,k]$. Use Karamata's Majorization Inequality as above or simply use Jensen Inequality.