Question

If \[{a^2} + {b^2} + {c^2} = 14\], then \[ab + bc + ca\] is always greater than or equal to?
A. 0
B. 14
C. -1
D. -7

Answer 1

Hint: A symmetric polynomial function is a polynomial where if you switch any pair of variables, it remains the same. The given equation is of the form \[{\left( {x + y + z} \right)^2}\] in which its expansion is \[{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx\] and according to this formula we can simplify the terms of \[ab + bc + ca\] by substituting \[{a^2} + {b^2} + {c^2} = 14\] in the obtained expression.

Complete step by step solution:
As per the given data: \[{a^2} + {b^2} + {c^2} = 14\]. Since square of any number cannot be negative, therefore, \[{\left( {a + b + c} \right)^2} \geqslant 0\] that is:
\[{\left( {a + b + c} \right)^2} \geqslant 0\]
Since, we know that
\[{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx\]
Expanding the terms, we get
\[ \Rightarrow \] \[{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca \geqslant 0\] ……………………. 1
As per the given data,
\[{a^2} + {b^2} + {c^2} = 14\]

Hence, applying this in equation 1 as
\[{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca \geqslant 0\]
\[\Rightarrow 14 + 2ab + 2bc + 2ca \geqslant 0\]
Combining the terms, we get
\[2\left( {ab + bc + ca} \right) \geqslant - 14\]
\[\Rightarrow ab + bc + ca \geqslant - \dfrac{{14}}{2}\]
\[ \Rightarrow \]\[ab + bc + ca \geqslant - 7\]
Hence,
\[ab + bc + ca \leqslant 7\]

Now consider,
\[{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \geqslant 0\]
As we know that
\[{\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy\]
Hence, the equation is
\[ \Rightarrow \] \[{a^2} + {b^2} - 2ab + {b^2} + {c^2} - 2bc + {c^2} + {a^2} - 2ca \geqslant 0\]
Simplifying the terms:
\[2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca \geqslant 0\]
\[\Rightarrow 2\left( {{a^2} + {b^2} + {c^2}} \right) - 2\left( {ab + bc + ca} \right) \geqslant 0\]
As per the given data,
\[{a^2} + {b^2} + {c^2} = 14\]
\[\Rightarrow 2\left( {14} \right) - 2\left( {ab + bc + ca} \right) \geqslant 0\]

Evaluate the terms as
\[28 - 2\left( {ab + bc + ca} \right) \geqslant 0\]
\[\Rightarrow - 2\left( {ab + bc + ca} \right) \geqslant - 28\]
\[\Rightarrow ab + bc + ca \geqslant \dfrac{{28}}{2}\]
Therefore, we get
\[ \Rightarrow \]\[ab + bc + ca \geqslant 14\]
Hence, if \[{a^2} + {b^2} + {c^2} = 14\], then \[ab + bc + ca\] is always greater than or equal to 14.

Hence, option B is the right answer.

Additional information:
A polynomial function is a function that can be expressed in the form of a polynomial. The definition can be derived from the definition of a polynomial equation. A polynomial is generally represented as \[P\left( x \right)\] and the highest power of the variable of \[P\left( x \right)\] is known as its degree. The domain of a polynomial function is entire real numbers (R).

Note: The key point to solve symmetric polynomial expressions is that when we solve the expression the resulting polynomial is the same as the initial polynomial and the square of every number is greater than or equal to 0 i.e., we know that \[{\left( {a + b + c} \right)^2} \geqslant 0\] and its expansion is \[{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca \geqslant 0\], hence expand the terms according to the equation and solve it.