# From Triangle Inequality to Inequality in Triangle

Let $a,$ $b,$ $c$ be the sides of a triangle. Introduce

$\begin{align} A&=(-a+ b+c)^{a}(a-b+c)^{b}(a+b-c)^{c};\\ B&=(-a+ b+c)^{a}(a-b+c)^{c}(a+b-c)^{b};\\ C&=(-a+ b+c)^{b}(a-b+c)^{a}(a+b-c)^{c};\\ D&=(-a+ b+c)^{b}(a-b+c)^{c}(a+b-c)^{a};\\ E&=(-a+ b+c)^{c}(a-b+c)^{a}(a+b-c)^{b};\\ F&=(-a+ b+c)^{c}(a-b+c)^{b}(a+b-c)^{a}.\\ \end{align}$

Prove that

$\max\{A,\;B,\;C,\;D,\;E,\;F\}\le a^{a}b^{b}c^{c}.$

The statement is due to Dorin Marghidanu and Kunihiko Chikaya. It was posted by Dorin at the CutTheKnotMath facebook page along with an observation that the inequality for $D$ has been established by Kunihiko Chikaya and a proof that established the inequality separately for all six quantities $A,\;B,\;C,\;D,\;E,\;F.$ A slight rewording of Dorin's proof highlights his approach.

### Generalization

Assume six positive quantities $x,\;y,\;z,\;a,\;b,\;c$ satisfy $x+y+z=a+b+c.$ Then

$x^{a}y^{b}z^{c}\le a^{a}b^{b}c^{c}.$

### Proof

The proof is a one-line application of the *weighted Arithmetic Mean - Geometric Mean inequality*:

Let positive $\lambda_{1},\ldots,\lambda_{n},$ $n$ a positive integer, satisfy $\displaystyle\sum_{k=1}^{n}\lambda_{k}=1.$ Then for positive $x_{1},\ldots,x_{n},$

$\lambda_{1}x_{1}+\ldots+\lambda_{n}x_{n}\ge x_{1}^{\lambda_{1}}\cdot\ldots\cdot x_{n}^{\lambda_{n}}.$

Note that

$\displaystyle\left[\left(\frac{x}{a}\right)^{a}\left(\frac{y}{b}\right)^{b}\left(\frac{z}{c}\right)^{c}\right]^{\frac{1}{a+b+c}}\le \frac{1}{a+b+c}\left(a\cdot\frac{x}{a}+b\cdot\frac{y}{b}+c\cdot\frac{z}{c}\right)=1.$

### Further generalization

Assume $\displaystyle\sum_{k=1}^{n}x_{k}=\sum_{k=1}^{n}a_{k},$ all quantities positive. Then

$x_{1}^{a_{1}}\cdot\ldots\cdot x_{n}^{a_{n}}\le a_{1}^{a_{1}}\cdot\ldots\cdot a_{n}^{a_{n}}.$

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