N. Paun, 1996

[Marius Mainea]

Aratati ca daca a,b,c sunt numere rationale pozitive, atunci numarul \( \sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\(\frac{a+b+c}{ab+bc+ca}\)^2} \) este rational.

GM 6\1994

[Claudiu Mindrila]

\( \sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\left(\frac{a+b+c}{ab+bc+ca}\right)^{2}}=\sqrt{\frac{\sum a^{2}b^{2}}{a^{2}b^{2}c^{2}}+\frac{\left(a+b+c\right)^{2}}{\left(ab+bc+ca\right)^{2}}}=\sqrt{\frac{\left(\sum a^{2}b^{2}\right)^{2}+2abc(a+b+c)\cdot\sum a^{2}b^{2}+a^{2}b^{2}c^{2}\left(a+b+c\right)^{2}}{a^{2}b^{2}c^{2}\left(ab+bc+ca\right)^{2}}}=
\sqrt{\frac{\left(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc\left(a+b+c\right)\right)^{2}}{a^{2}b^{2}c^{2}\left(ab+bc+ca\right)^{2}}} \).
Deci
\( LHS=\frac{\left(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc\left(a+b+c\right)\right)}{abc\left(ab+bc+ca\right)}=\frac{1}{ab+bc+ca}\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)+2\cdot\frac{a+b+c}{ab+bc+ca}\in\mathbb{Q} \)

[mihai++]

e fara 2 acolo sub patrat