
If a, b, c are the sides of a triangle, then the minimum value of \[\dfrac{a}{{b + c - a}} + \dfrac{b}{{c + a - b}} + \dfrac{c}{{a + b - c}}\]is equal to
A. 3
B. 6
C. 9
D. 12
Answer
216.3k+ views
Hint: In order to find the minimum value of\[\dfrac{a}{{b + c - a}} + \dfrac{b}{{c + a - b}} + \dfrac{c}{{a + b - c}}\], all you need to do is apply distributive property and solve for each variable separately.
For example, let's say we have 3 variables (A, B, C) and we want to find out what their combined minimum value would be.
Complete step by step solution:\[a,{\rm{ }}b,{\rm{ }}c > 0\]
Take \[\dfrac{a}{{b + c - a}} + \dfrac{b}{{c + a - b}} + \dfrac{c}{{a + b - c}}\]
\[2A = \dfrac{{2a}}{{b + c - a}} + \dfrac{{2b}}{{c + a - b}} + \dfrac{{2c}}{{a + b - c}}\]
To each term, add 1 and subtract 3.
\[2A = \left[ {\dfrac{{2a}}{{b + c - a}} + 1} \right] + \left[ {\dfrac{{2b}}{{c + a - b}} + 1} \right] + \left[ {\dfrac{{2c}}{{a + b - c}} + 1} \right] - 3\]
\[2A = \left[ {\dfrac{{a + b + c}}{{b + c - a}}} \right] + \left[ {\dfrac{{a + b + c}}{{c + a - b}}} \right] + \left[ {\dfrac{{a + b + c}}{{a + b - c}}} \right] - 3\]
\[ = \left( {a + b + c} \right) + \left[ {\dfrac{1}{{b + c - a}} + \dfrac{1}{{c + a - b}} + \dfrac{1}{{a + b - c}}} \right] - 3\]
For the numbers, we use A.M. > H.M
\[\dfrac{1}{{b + c - a}},\dfrac{1}{{a + c - b}},\dfrac{1}{{a + b - c}}\]
\[\dfrac{{\dfrac{1}{{b + c - a}} + \dfrac{1}{{a + c - b}} + \dfrac{1}{{a + b - c}}}}{3} > \dfrac{3}{{(b + c - a) + (a + c - b) + (a + b - c)}}\]
\[ > \dfrac{3}{{a + b + c}}\]
\[(a + b + c)\left( {\dfrac{1}{{b + c - a}} + \dfrac{1}{{a + c - b}} + \dfrac{1}{{a + b - c}}} \right) > g\]
\[2A + 3 > g\]
\[2A{\rm{ }} + {\rm{ }}3{\rm{ }} > {\rm{ }}9\]
\[2A{\rm{ }} > {\rm{ }}6\]
\[A{\rm{ }} > {\rm{ }}3\]
Therefore, \[\dfrac{a}{{b + c - a}} + \dfrac{b}{{c + a - b}} + \dfrac{c}{{a + b - c}}\]must equal at least that value. equivalent to \[3\].
Option ‘A’ is correct
Note: Students often make mistakes when using A.M> H.M. This is because the order of operations (or the algebraic rules) do not always work in reverse.
The maximum value of \[\dfrac{a}{{b + c - a}} + \dfrac{b}{{c + a - b}} + \dfrac{c}{{a + b - c}}\]is equal to 3. So, the students make mistake by thinking that it will be a maximum when in fact it would always be the same as (3).
For example, let's say we have 3 variables (A, B, C) and we want to find out what their combined minimum value would be.
Complete step by step solution:\[a,{\rm{ }}b,{\rm{ }}c > 0\]
Take \[\dfrac{a}{{b + c - a}} + \dfrac{b}{{c + a - b}} + \dfrac{c}{{a + b - c}}\]
\[2A = \dfrac{{2a}}{{b + c - a}} + \dfrac{{2b}}{{c + a - b}} + \dfrac{{2c}}{{a + b - c}}\]
To each term, add 1 and subtract 3.
\[2A = \left[ {\dfrac{{2a}}{{b + c - a}} + 1} \right] + \left[ {\dfrac{{2b}}{{c + a - b}} + 1} \right] + \left[ {\dfrac{{2c}}{{a + b - c}} + 1} \right] - 3\]
\[2A = \left[ {\dfrac{{a + b + c}}{{b + c - a}}} \right] + \left[ {\dfrac{{a + b + c}}{{c + a - b}}} \right] + \left[ {\dfrac{{a + b + c}}{{a + b - c}}} \right] - 3\]
\[ = \left( {a + b + c} \right) + \left[ {\dfrac{1}{{b + c - a}} + \dfrac{1}{{c + a - b}} + \dfrac{1}{{a + b - c}}} \right] - 3\]
For the numbers, we use A.M. > H.M
\[\dfrac{1}{{b + c - a}},\dfrac{1}{{a + c - b}},\dfrac{1}{{a + b - c}}\]
\[\dfrac{{\dfrac{1}{{b + c - a}} + \dfrac{1}{{a + c - b}} + \dfrac{1}{{a + b - c}}}}{3} > \dfrac{3}{{(b + c - a) + (a + c - b) + (a + b - c)}}\]
\[ > \dfrac{3}{{a + b + c}}\]
\[(a + b + c)\left( {\dfrac{1}{{b + c - a}} + \dfrac{1}{{a + c - b}} + \dfrac{1}{{a + b - c}}} \right) > g\]
\[2A + 3 > g\]
\[2A{\rm{ }} + {\rm{ }}3{\rm{ }} > {\rm{ }}9\]
\[2A{\rm{ }} > {\rm{ }}6\]
\[A{\rm{ }} > {\rm{ }}3\]
Therefore, \[\dfrac{a}{{b + c - a}} + \dfrac{b}{{c + a - b}} + \dfrac{c}{{a + b - c}}\]must equal at least that value. equivalent to \[3\].
Option ‘A’ is correct
Note: Students often make mistakes when using A.M> H.M. This is because the order of operations (or the algebraic rules) do not always work in reverse.
The maximum value of \[\dfrac{a}{{b + c - a}} + \dfrac{b}{{c + a - b}} + \dfrac{c}{{a + b - c}}\]is equal to 3. So, the students make mistake by thinking that it will be a maximum when in fact it would always be the same as (3).
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