cho M = (a^2+b^2-c^2)/2ab + (b^2+c^2-a^2)/2bc + c^2+a^2-b^2/2ca Với a,b,c là độ dài 3 cạnh tam giác, chứng minh M>1

Giải thích các bước giải:

Ta có:

$M=\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{b^2+c^2-a^2}{2bc}+\dfrac{c^2+a^2-b^2}{2ca}$

$\to M=\dfrac{c(a^2+b^2-c^2)+a(b^2+c^2-a^2)+b(c^2+a^2-b^2)}{2abc}$

$\to M=\dfrac{ca^2+cb^2-c^3+ab^2+c^2a-a^3+c^2b+a^2b-b^3}{2abc}$

$\to M=\dfrac{(ab^2+cb^2-b^3)+(ca^2+c^2a-abc)-(c^3+a^3)+(c^2b+a^2b-abc)+2abc}{2abc}$

$\to M=\dfrac{b^2(a+c-b)+ac(a+c-b)-(a+c)(a^2-ac+c^2)+b(c^2+a^2-ac)+2abc}{2abc}$

$\to M=\dfrac{b^2(a+c-b)+ac(a+c-b)-(a+c-b)(a^2-ac+c^2)+2abc}{2abc}$

$\to M=\dfrac{(a+c-b)(b^2+ac-(a^2-ac+c^2))+2abc}{2abc}$

$\to M=\dfrac{(a+c-b)(b^2+ac-a^2+ac-c^2)+2abc}{2abc}$

$\to M=\dfrac{(a+c-b)(b^2-a^2+2ac-c^2)+2abc}{2abc}$

$\to M=\dfrac{(a+c-b)(b^2-(a^2-2ac+c^2))+2abc}{2abc}$

$\to M=\dfrac{(a+c-b)(b^2-(a-c)^2)+2abc}{2abc}$

$\to M=\dfrac{(a+c-b)(b+a-c)(b-a+c)+2abc}{2abc}$

$\to M=\dfrac{(a+c-b)(b+a-c)(b-a+c)}{2abc}+1$

Mà $a,b,c$ là cạnh của $1$ tam giác

$\to a+c>b, b+a>c, b+c>a$

$\to a+c-b>0, b+a-c>0, b-a+c>0$

$\to M>0+1$

$\to M>1$