"An inequality with cos⁡ and triangle sidesHere is..." solved and explained

chechemuaen7

Answered question

2022-03-16

An inequality with

\cos

and triangle sides
Here is the problem:
Let ABC be a triangle with sides a,b,c. Show that

\frac{\cos A}{a^{3}} + \frac{\cos B}{b^{3}} + \frac{\cos C}{c^{3}} \geq \frac{3}{2 a b c}

Roland Ramsey

Beginner2022-03-17Added 4 answers

By the cosine formula, we have

\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}

etc, which the left hand side can be transformed into:

\frac{a^{2} + b^{2} - c^{2}}{2 a b c^{3}} + \frac{a^{2} + c^{2} - b^{2}}{2 a b^{3} c} + \frac{b^{2} + c^{2} - a^{2}}{2 a^{3} b c}

If you factor out

\frac{1}{2 a b c}

from the expression you obtained, you get

\frac{1}{2 a b c} (\frac{a^{2} + b^{2} - c^{2}}{c^{2}} + \frac{a^{2} + c^{2} - b^{2}}{b^{2}} + \frac{b^{2} + c^{2} - a^{2}}{a^{2}})

So you just need to prove

\frac{a^{2} + b^{2} - c^{2}}{c^{2}} + \frac{a^{2} + c^{2} - b^{2}}{b^{2}} + \frac{b^{2} + c^{2} - a^{2}}{a^{2}} \geq 3

. Writting

\frac{a^{2} + b^{2} - c^{2}}{c^{2}} + \frac{a^{2} + c^{2} - b^{2}}{b^{2}} + \frac{b^{2} + c^{2} - a^{2}}{a^{2}}

= (\frac{a^{2}}{b^{2}} + \frac{b^{2}}{a^{2}}) + (\frac{a^{2}}{c^{2}} + \frac{c^{2}}{a^{2}}) + (\frac{b^{2}}{c^{2}} + \frac{c^{2}}{b^{2}}) - 3

it is enough to show that each of the terms in the parentheses is at least 2.

pintorreeqwf

Beginner2022-03-18Added 3 answers

2 a b c \sum_{c y c} \frac{a^{2} + b^{2} - c^{2}}{2 a b c^{3}} =

= \sum_{c y c} (\frac{a^{2}}{c^{2}} + \frac{b^{2}}{c^{2}} - 1) =

= - 3 + \sum_{c y c} (\frac{a^{2}}{b^{2}} + \frac{b^{2}}{a^{2}}) =

= - 3 + \sum_{c y c} (2 + {(\frac{a}{b} - \frac{b}{a})}^{2}) =

= + 3 + \sum_{c y c} {(\frac{a}{b} - \frac{b}{a})}^{2} \geq 3

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