SolFeb2205

Solution by landen

Show that if $c,b,a > 0\,$ then

$c^c\, b^b \,a^a \ge (c\,b\,a)^{\frac{c+b+a}{3}}$

It looks like logs will help

$\log \left(c^{c}\,b^{b}\,a^{a}\right)=c\,\log c+b\,\log b+a\,\log a$

Define: $f\left(x\right)=x\,\log x$

${{d^2}\over{d\,x^2}}\,\left(x\,\log x\right)={{1}\over{x}}>0$

Since the second derivative is positive, $f(x)\,$ is convex. For a convex function we can apply Jensen's (http://www.engineering.usu.edu/classes/ece/7680/lecture2/node5.html) inequality, the average of the function at points is greater than or equal to the function evaluated at the average of the points:

$\frac{f(c)+f(b)+f(a)}{3}\ge f\left(\frac{c+b+a}{3}\right)$

${{c\,\log c+b\,\log b+a\,\log a}\over{3}}\geq {{\left(c+b+a\right) \,\log \left({{c+b+a}\over{3}}\right)}\over{3}}$

Since $\log(x)\,$ is monotonically increasing we can use the arithmetic-geometric mean inequality (http://www.cs.utexas.edu/users/misra/Notes.dir/Arithmetic,GeometricMean.pdf):

$\log \left({{c+b+a}\over{3}}\right)\geq \log \left(c^{{{1}\over{3}} }\,b^{{{1}\over{3}}}\,a^{{{1}\over{3}}}\right)=\frac{1}{3}\log(c\,b\,a)$

$c\,\log c+b\,\log b+a\,\log a\geq {{\left(c+b+a\right)\,\log \left( c\,b\,a\right)}\over{3}}$

Since $e^x\,$ is monotonically increasing we can exponentiate both sides to get:

$c^c\, b^b \,a^a \ge (c\,b\,a)^{\frac{c+b+a}{3}}$

Q.E.D.

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