If \[a,b,c\] are in descending order. Prove that \[{\left( {\dfrac{{a + c}}{{a - c}}} \right)^a} < {\left( {\dfrac{{b + c}}{{b - c}}} \right)^b}\]
VerifiedHint: Here we use the concept of descending order which means as we go up the scale value decreases, so each element is greater than the next element. Also changing inequalities by adding, subtracting, multiplying and dividing a positive number helps us in comparing the inequalities.
* The descending numbers \[a,b,c\] can be represented on the real line.
Complete step by step answer:
We know \[a,b,c\] are in descending order, therefore \[a > b > c\]
Let us take \[a > b\]
Adding \[c\] on both sides of inequality we get
\[a + c > b + c\] \[...(i)\]
Similarly, subtracting \[c\] from both sides of inequality we get
\[a - c > b - c\]
Since, taking reciprocal on both sides changes the inequality
Therefore, \[\dfrac{1}{{a - c}} < \dfrac{1}{{b - c}}\] \[...(ii)\]
Multiplying equations \[(i)\] and \[(ii)\].
\[\dfrac{{a + c}}{{a - c}} < \dfrac{{b + c}}{{b - c}}\] \[...(iii)\]
Taking an example to prove the above statement, let \[a = 3,b = 2,c = 1\] .
From equation \[(i)\]
\[
a + c > b + c \\
3 + 1 > 2 + 1 \\
4 > 3 \\
\]
Which holds true .
Now , from equation \[(ii)\]
\[
\dfrac{1}{{a - c}} < \dfrac{1}{{b - c}} \\
\dfrac{1}{{3 - 1}} < \dfrac{1}{{2 - 1}} \\
\dfrac{1}{2} < 1 \\
\]
Which holds true .
From equation \[(iii)\]
\[
\dfrac{{a + c}}{{a - c}} < \dfrac{{b + c}}{{b - c}} \\
\dfrac{{3 + 1}}{{3 - 1}} < \dfrac{{2 + 1}}{{2 - 1}} \\
\dfrac{4}{2} < \dfrac{3}{1} \\
2 < 3 \\
\]
Therefore, the statement holds true.
If a number is less than any other number , taking greater power on the lesser side will make the denominator more and more therefore the whole number becomes less.
Since, \[\dfrac{{a + c}}{{a - c}} < \dfrac{{b + c}}{{b - c}}\]
Therefore taking the greater power on LHS of the equation , the denominator increases and so the whole LHS of the equation decreases.
As\[a > b\], therefore taking power \[a\] on LHS and power \[b\] on RHS of the equation.
Therefore, \[{\left( {\dfrac{{a + c}}{{a - c}}} \right)^a} < {\left( {\dfrac{{b + c}}{{b - c}}} \right)^b}\]
Note:
In these types of questions students are likely to make mistakes while multiplying the inequalities. Students should know the difference between ascending order( moving up the scale) and descending order(moving down the scale) to solve such questions.
Some inequality changing rules to keep in mind are
1) If \[a < b\] then \[a + c < b + c\]
2) If \[a < b\] then \[a - c < b - c\]
3) If \[a < b\] and \[c\] is a positive number then \[a.c < b.c\]
4) If \[a < b\] and \[c\] is a negative number then \[a.c > b.c\]
