Sunday, November 22, 2009

Triangle Inequality

Given $n$ right triangles each with sides $a_i, b_i, c_i$, with $c_i$ being the hypotenuse ($i=1,\cdots, n)$. Let $A = \sum a_i, B = \sum b_i, C= \sum c_i$.

Prove that

$\displaystyle \frac{a_1b_1}{c_1} + \cdots + \frac{a_nb_n}{c_n} \leq \frac{AB}{\sqrt{A^2+B^2}}$

Harder version: prove that

$\displaystyle \frac{a_1b_1}{c_1} + \cdots + \frac{a_nb_n}{c_n} \leq \frac{AB}{C}$


For each $i$,

$ (\frac{1}{a_i^2} + \frac{1}{b_i^2})(\frac{1}{A^2} + \frac{1}{B^2}) \geq (\frac{1}{a_iA} + \frac{1}{b_iB})^2$


$ (\frac{a_i}{A^3} + \frac{b_i}{B^3})(\frac{1}{a_iA} + \frac{1}{b_iB}) \geq (\frac{1}{A^2} + \frac{1}{B^2})^2$

If we let $S = 1/A^2 + 1/B^2$, then two inequalities above can be combined to form:

$ (\frac{a_i}{A^3} + \frac{b_i}{B^3})\sqrt{\frac{1}{a_i^2} + \frac{1}{b_i^2}} \geq S^{3/2}$

$ (\frac{a_i}{A^3} + \frac{b_i}{B^3}) \geq S^{3/2} \frac{a_ib_i}{c_i}$

So, when we sum the last inequality over all $i$, we have:

$ (\frac{A}{A^3} + \frac{B}{B^3}) \geq S^{3/2} \sum \frac{a_ib_i}{c_i}$

$ S^{-1/2} \geq \sum \frac{a_ib_i}{c_i}$

So $LHS \leq \frac{1}{\sqrt{S}} =RHS$

The harder version is stronger than the easier version due to Minkowski's inequality, which asserts that $C^2 \geq A^2+B^2$
Solution for harder version

We wish to prove that

$ \sum_i c_i \sum_j \frac{a_jb_j}{c_j} \leq \sum_i a_i \sum_j b_j$

Consider the expansion for both sides. If $i=j$ then the product is $a_ib_i$, which is canceled in both LHS and RHS.

If $i \neq j$, then we assert that:

$ \frac{c_ia_jb_j}{c_j} +\frac{c_ja_ib_i}{c_i} \leq a_ib_j + a_jb_i$

$ \iff c_i^2a_jb_j +c_j^2a_ib_i \leq (a_ib_j + a_jb_i)c_ic_j$

$ \iff (a_i^2+b_i^2)a_jb_j +(a_j^2+b_j^2)a_ib_i \leq (a_ib_j + a_jb_i)c_ic_j$

$ \iff (a_ib_j + a_jb_i)(a_ia_j + b_ib_j) \leq (a_ib_j + a_jb_i)c_ic_j$

$ \iff a_ia_j + b_ib_j \leq c_ic_j$

which is true by Cauchy, since $c_i = \sqrt{a_i^2+b_i^2}$

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