Question

If abcd = 1, where a, b, c, d are positive reals then the minimum value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+ab+bc+cd+ad+ac+bd$
( a ) 6
( b ) 10
( c ) 12
( d ) 20

Answer 1

Hint: For, solving this question we will use concept of A.M and G.M and the relation between A.M and G.M which is given as A.M of these n items is always equals to or greater than G.M of these n items that is $\text{A}\text{.M}\ge \text{G}\text{.M}$ by which we will get the possible minimum value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+ab+bc+cd+ad+ac+bd$.

Complete step-by-step solution:
In question, it is given that abcd = 1 and all four numbers a, b, c, d are positive real numbers.
We have to find the minimum possible value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+ab+bc+cd+ad+ac+bd$.
Okay, now we will first see what is Arithmetic Mean ( A.M ) and geometric Mean ( G.M )
The arithmetic mean is the average of a set of numerical values, as calculated by adding them together and dividing them by the number of terms in the set.
So, if we have n items say, ${{n}_{1}},{{n}_{2}},.......,{{n}_{n}}$ then, A.M of n items will be equals to
$\text{A}\text{.M = }\dfrac{{{n}_{1}}+{{n}_{2}}+.......+{{n}_{n}}}{n}$
Geometric mean is the calculation by first doing product of n terms and then finding ${{\text{n}}^{\text{th}}}$ root of product.
So, if we have n items say, ${{n}_{1}},{{n}_{2}},.......,{{n}_{n}}$ then, A.M of n items will be equals to
$\text{G}\text{.M = (}{{n}_{1}}\cdot {{n}_{2}}\cdot .......\cdot {{n}_{n}}{{)}^{\dfrac{1}{n}}}$
Now, let we have 10 terms as \[{{\text{a}}^{\text{2}}}\text{, }{{\text{b}}^{\text{2}}}\text{, }{{\text{c}}^{\text{2}}}\text{, }{{\text{d}}^{\text{2}}}\text{, ab, bc, cd, ad, ac, bd}\], the A.M and G.M of numbers will be
\[\text{A}\text{.M = }\dfrac{\text{(}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}{{\text{d}}^{\text{2}}}\text{+ab+bc+cd+ad+ac+bd})}{10}\]
\[\text{G}\text{.M = (}{{\text{a}}^{\text{2}}}\cdot {{\text{b}}^{\text{2}}}\cdot {{\text{c}}^{\text{2}}}\cdot {{\text{d}}^{\text{2}}}\cdot \text{ab}\cdot \text{bc}\cdot \text{cd}\cdot \text{ad}\cdot \text{ac}\cdot \text{bd}{{)}^{\dfrac{1}{10}}}\]
Now, we know that if we have n items say ${{n}_{1}},{{n}_{2}},.......,{{n}_{n}}$, then A.M of these n items is always equals to or greater than G.M of these n items that is
$\text{A}\text{.M}\ge \text{G}\text{.M}$
So, for 10 numbers \[{{\text{a}}^{\text{2}}}\text{, }{{\text{b}}^{\text{2}}}\text{, }{{\text{c}}^{\text{2}}}\text{, }{{\text{d}}^{\text{2}}}\text{, ab, bc, cd, ad, ac, bd}\],
\[\dfrac{\text{ (}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}{{\text{d}}^{\text{2}}}\text{+ab+bc+cd+ad+ac+bd})}{10}\ge {{\text{(}{{\text{a}}^{\text{2}}}\cdot {{\text{b}}^{\text{2}}}\cdot {{\text{c}}^{\text{2}}}\cdot {{\text{d}}^{\text{2}}}\cdot \text{ab}\cdot \text{bc}\cdot \text{cd}\cdot \text{ad}\cdot \text{ac}\cdot \text{bd})}^{\dfrac{1}{10}}}\]
On simplifying we get,
\[\dfrac{\text{ (}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}{{\text{d}}^{\text{2}}}\text{+ab+bc+cd+ad+ac+bd})}{10}\ge {{\text{(}{{\text{a}}^{5}}\cdot {{\text{b}}^{5}}\cdot {{\text{c}}^{5}}\cdot {{\text{d}}^{5}})}^{\dfrac{1}{10}}}\]
\[\dfrac{\text{ (}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}{{\text{d}}^{\text{2}}}\text{+ab+bc+cd+ad+ac+bd})}{10}\ge {{\text{(a}\cdot \text{b}\cdot \text{c}\cdot \text{d})}^{5\times }}^{\dfrac{1}{10}}\]
On simplifying, we get
\[\dfrac{\text{ (}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}{{\text{d}}^{\text{2}}}\text{+ab+bc+cd+ad+ac+bd})}{10}\ge {{\text{(a}\cdot \text{b}\cdot \text{c}\cdot \text{d})}^{\dfrac{1}{2}}}\]
As it is given that, abcd = 1
So, \[\dfrac{\text{ (}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}{{\text{d}}^{\text{2}}}\text{+ab+bc+cd+ad+ac+bd})}{10}\ge {{\text{(1})}^{\dfrac{1}{2}}}\]
\[\dfrac{\text{ (}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}{{\text{d}}^{\text{2}}}\text{+ab+bc+cd+ad+ac+bd})}{10}\ge 1\]
Taking 10 from denominator on left hand side to numerator on right hand side by using cross multiplication, we get
\[\text{ (}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}{{\text{d}}^{\text{2}}}\text{+ab+bc+cd+ad+ac+bd})\ge 10\]
Hence, we get \[\text{ (}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}{{\text{d}}^{\text{2}}}\text{+ab+bc+cd+ad+ac+bd})\ge 10\] which means value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+ab+bc+cd+ad+ac+bd$is always equals to or greater than 10.
So, minimum value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+ab+bc+cd+ad+ac+bd$ is 10. Hence, option ( b ) is correct.

Note: For, solving such question we can use concept of A.M and G.M so, one must know the meaning, formula and relation of A.M and G.M that is, if we have n items say, ${{n}_{1}},{{n}_{2}},.......,{{n}_{n}}$ then, $\text{A}\text{.M = }\dfrac{{{n}_{1}}+{{n}_{2}}+.......+{{n}_{n}}}{n}$ and $\text{G}\text{.M = (}{{n}_{1}}\cdot {{n}_{2}}\cdot .......\cdot {{n}_{n}}{{)}^{\dfrac{1}{n}}}$. It is always important to solve questions carefully with no error as answer may get change we can make equation more complex.