inequality and sides of triangles (1 Viewer)

Mahan1 · Feb 19, 2017

If alb,c are sides a triangle, then prove

$3(ab+bc+ca) \leq (a+b+c)^{2} \leq 4(ab+bc+ca)$

Mahan1 · Feb 27, 2017

This is what I got:
the first inequality:

$(b-c)^{2} \geq 0 , (a-c)^{2} \geq 0 , (a-b)^{2} \geq 0$

adding all of them together we get:

$2(a^{2} + b^{2} +c^{2}) \geq 2(ab + ac +bc)$

$(a^{2} + b^{2} +c^{2}) \geq (ab + ac +bc)$

therefore,

$(a+b+c)^{2} \geq 3(ab + ac +bc)$

the second inequality:

by triangle inequality we have:

$a > |b-c|, b>|a-c|, c> |a-b|$

by squaring both sides, we get :

$a^{2} > c^{2} + b^{2} -2bc$

$b^{2} > c^{2} + a^{2} -2ac$

$c^{2} > a^{2} + b^{2} -2ba$

adding them up:

$a^{2} + b^{2} + c^{2} > 2(a^{2} + b^{2} + c^{2})-2(ab+ac+bc)$

$2ab+2ac+2bc > a^{2} + b^{2} + c^{2}$

by adding $2ab+2ac+2bc$

we get,

$4(ab+ac+bc) > (a+b+c)^{2}$

inequality and sides of triangles (1 Viewer)

Mahan1

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Mahan1

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