Question

Cho a,b,c \(\in\) Q thỏa mãn a.b+b.c+c.a =1

CM:\(\sqrt{\left(a^2+1\right).\left(b^2+1\right).\left(c^2+1\right)}\in Q\)

Answer 1

thay trực tiếp giả thiết ta có

\(\sqrt{\left(a^2+1\right)}=\sqrt{a^2+ab+bc+ac}=\sqrt{a\left(a+b\right)+c\left(a+b\right)}=\sqrt{\left(a+c\right)\left(a+b\right)}\)

tương tự ta có

\(\sqrt{b^2+1}=\sqrt{\left(b+a\right)\left(b+c\right)}\)

\(\sqrt{c^2+1}=\sqrt{\left(c+a\right)\left(c+b\right)}\)

nên

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}=\sqrt{\left(\left(a+b\right)\left(a+c\right)\left(b+c\right)\right)^2}=\left|\left(a+b\right)\left(a+c\right)\left(b+c\right)\right|\)

mà \(a,b,c\in Q\) nên \(\left|\left(a+b\right)\left(a+c\right)\left(b+c\right)\right|\in Q\Rightarrowđpcm\)