Question

Sa se arate că dacă numerele a, b, c sunt în progresie geometrică, atunci:
a³+b³+c³=a²b²c²(1/a³+1/b³+1/c³)

Dau coronița!

Answer 1

$\displaystyle\sf\\numerele~a,b~si~c~sunt~in~progresie~geometrica,~ceea~ce~implica~ca\\\boxed{\sf b=\sqrt{ac}}.\\---------------------\\a^3+b^3+c^3=a^2b^2c^2\bigg(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\bigg) \Leftrightarrow \frac{a^3+b^3+c^3}{a^2b^2c^2}=\bigg(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\bigg) \Leftrightarrow\\\frac{a^3+ac\sqrt{ac}+c^3}{a^3c^3}=\frac{a^3b^3+b^3c^3+a^3c^3}{a^3b^3c^3} =\frac{a^4c\sqrt{ac}+c^4a\sqrt{ac}+a^3c^3}{a^4c^4\sqrt{ac}}=$

$\displaystyle\sf\\\frac{ac(a^3\sqrt{ac}+c^3\sqrt{ac}+a^2c^2}{(ac)^4\sqrt{ac}}=\frac{a^3\sqrt{ac}+c^3\sqrt{ac}+a^2c^2}{a^3c^3\sqrt{ac}},~rationalizam \implies \\obtinem:\frac{a^3+c^3+ac\sqrt{ac}}{a^3c^3},~observam~ca~avem~egalitate.\\Prin~urmare,~avem~aceassta~egalitate~daca~a,b,c~sunt~in~progresie~geometrica.$