Question

a^2/b^2 + b^2/c^2 + c^2/a^2= a/c+ b/a + c/b demonstrati ca a=b=c

Answer 1

[tex] \frac{a^2}{b^2}+ \frac{b^2}{c^2} + \frac{c^2}{a^2}= \frac{a}{c}+ \frac{b}{a}+ \frac{c}{b} \\ \frac{a^4c^2 + a^2b^4 + c^4b^2}{a^2b^2c^2} = \frac{a^2b+b^2c + c^2a}{abc} \\ \frac{a^4c^2 + a^2b^4 + c^4b^2}{abc} = a^2b+b^2c + c^2a \\ \frac{a^4c^2 + a^2b^4 + c^4b^2}{abc} = \frac{abc(a^2b+b^2c + c^2a)}{abc} \\ \frac{a^4c^2 + a^2b^4 + c^4b^2}{abc} = \frac{a^3b^2c+ab^3c^2 + bc^3a^2}{abc} \\ a^4c^2 + a^2b^4 + c^4b^2 = a^3b^2c+ab^3c^2 + bc^3a^2 \\a^4c^2 + a^2b^4 + c^4b^2 - a^3b^2c-ab^3c^2 - bc^3a^2 = 0 \\ ac^2(a^3-b^3) + ba^2(b^3 - c^3) + cb^2(c^3-a^3) = 0 \\[/tex]
Pentru ca ecuatia sa aiba loc, trebuie ca fiecare din cei trei termeni sa fie 0. Astfel, ori $ac^2, ba^2, cb^2$ sunt 0 (caz in care a = b = c = 0 - ceea ce nu se poate, deoarece sunt numitori => caz imposibil), ori (a^3-b^3), (b^3 - c^3), (c^3-a^3) sunt egale, caz in care obligatoriu a = b = c. Sper ca ti-am fost de folos.