Question:
If \[ a+b+c=6 \] and \[ \frac1a+\frac1b+\frac1c=\frac32 \] find:
\[ \frac ab+\frac ac+\frac ba+\frac bc+\frac ca+\frac cb \]
Solution:
Given:
\[ \frac1a+\frac1b+\frac1c=\frac32 \]
Taking LCM:
\[ \frac{ab+bc+ca}{abc}=\frac32 \]
Using identity:
\[ (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \]
Now,
\[ \frac ab+\frac ac+\frac ba+\frac bc+\frac ca+\frac cb \]
\[ = \frac{a^2+b^2+c^2}{ab+bc+ca} \]
Also,
\[ a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca) \]
\[ =6^2-2(ab+bc+ca) \]
\[ =36-2(ab+bc+ca) \]
From \[ \frac{ab+bc+ca}{abc}=\frac32 \] we get:
\[ 2(ab+bc+ca)=3abc \]
Since \[ a+b+c=6 \] and \[ 2(ab+bc+ca)=3abc \]
Using identity:
\[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \]
For \[ a=b=c=2 \] the conditions are satisfied.
Therefore,
\[ \frac ab+\frac ac+\frac ba+\frac bc+\frac ca+\frac cb \]
\[ = 1+1+1+1+1+1 \]
\[ =6 \]