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Inequality 38(Christos Patilas)

Problem:

If $\displaystyle a,b,c$ are angles of an acute triangle, prove that $\displaystyle \pi^{\pi}a^bb^cc^a\leq (a^2+b^2+c^2)^{\pi}$.

Solution:

The function $\displaystyle f(x)=\ln x$ is strictly concave, so from the general weighted Jensen inequality with weights the $\displaystyle a,b,c$ we have that

$\displaystyle \frac{b \ln a+c \ln b+a \ln c}{a+b+c}\leq \ln\left(\frac{ab+bc+ca}{a+b+c}\right)$, or $\displaystyle \frac{b \ln a+c \ln b+a \ln c}{\pi}\leq \ln\left(\frac{ab+bc+ca}{\pi}\right)$.

But, the last relation can be rewritten as

$\displaystyle \frac{1}{\pi}\cdot \ln(a^bb^cc^a)\leq \ln\left(\frac{ab+bc+ca}{\pi}\right)$, that is $\displaystyle \ln\left(a^bb^cc^a\right)^{1/\pi}\leq \ln \left(\frac{ab+bc+ca}{\pi}\right)$.

Removing the logarithm we get

$\displaystyle \left(a^bb^cc^a\right)^{1/\pi}\leq\frac{ab+bc+ca}{\pi}\Longrightarrow \pi^{\pi}a^bb^cc^a\leq \left(ab+bc+ca\right)^{\pi}\leq (a^2+b^2+c^2)^{\pi}$, Q.E.D.