Question

If \[x > 0\], \[y > 0\] and \[{x^2} + {y^2} = 8\] \[\left[ {x \in R} \right]\], then
A) \[x + y \geqslant 4\]
B) \[x + y \leqslant 4\]
C) \[x + y = 1\]
D) None of these

Answer 1

Hint: Here, we will first use one of the properties of arithmetic mean that is
\[\dfrac{{\left( {{a^n} + {b^n}} \right)}}{2} \leqslant {\left( {\dfrac{{a + b}}{2}} \right)^n}\] if \[{\text{0}} \leqslant {\text{n}} \leqslant {\text{1}}\]
Then in this equation we will put the value of \[a\] as \[{x^2}\], \[b\] as \[{y^2}\] and \[n = \dfrac{1}{2}\] . Also, it is given in the question that \[{x^2} + {y^2} = 8\], so we will put this value in the right-hand side of the above equation to find the required relation between \[a\] and \[b\].

Complete step by step answer:
By the properties of arithmetic mean we can write
\[\left( {\dfrac{{{a^n} + {b^n}}}{2}} \right) \leqslant {\left( {\dfrac{{a + b}}{2}} \right)^n} - - - (1)\] if \[{\text{0}} \leqslant {\text{n}} \leqslant {\text{1}}\]
Putting \[a = {x^2}\], \[b = {y^2}\] and \[n = \dfrac{1}{2}\] in equation \[(1)\] it becomes
\[\left( {\dfrac{{{{\left( {{x^2}} \right)}^{\dfrac{1}{2}}} + {{\left( {{y^2}} \right)}^{\dfrac{1}{2}}}}}{2}} \right) \leqslant {\left( {\dfrac{{{x^2} + {y^2}}}{2}} \right)^{\dfrac{1}{2}}}\]
Now in the right-hand side of the above equation we will put the value of \[{x^2} + {y^2} = 8\] , which is given in the question. Now the equation becomes;
\[ \Rightarrow \;\left( {\dfrac{{{{\left( {{x^2}} \right)}^{\dfrac{1}{2}}} + {{\left( {{y^2}} \right)}^{\dfrac{1}{2}}}}}{2}} \right) \leqslant {\left( {\dfrac{8}{2}} \right)^{\dfrac{1}{2}}}\]
On shifting \[2\] to the right-hand side, we get
\[ \Rightarrow \;{\left( {{x^2}} \right)^{\dfrac{1}{2}}} + {\left( {{y^2}} \right)^{\dfrac{1}{2}}} \leqslant 2 \times {\left( {\dfrac{8}{2}} \right)^{\dfrac{1}{2}}}\]
On simplification the above equation becomes
\[ \Rightarrow \;{\left( {{x^2}} \right)^{\dfrac{1}{2}}} + {\left( {{y^2}} \right)^{\dfrac{1}{2}}} \leqslant 2 \times {\left( 4 \right)^{\dfrac{1}{2}}}\]
On further simplifying the right-hand side, we get
\[ \Rightarrow \;{\left( {{x^2}} \right)^{\dfrac{1}{2}}} + {\left( {{y^2}} \right)^{\dfrac{1}{2}}} \leqslant 2 \times {\left( {{{(2)}^2}} \right)^{\dfrac{1}{2}}}\]
On simplifying both the sides, we get
\[ \Rightarrow x + y \leqslant 2 \times 2\]
\[ \Rightarrow x + y \leqslant 4\]
Therefore, \[x + y \leqslant 4\].
Hence, option (B) is correct.

Additional information:
To check whether this property is true or not we can use mathematical induction or for our convenience we can even check by just putting random values of the variables. Let us take some random values of \[a,b,n\].
Let,
\[a = 8\]
\[b = 27\]
\[n = \dfrac{1}{3}\]
So, the left-hand side of the above equation becomes
\[\left( {\dfrac{{{a^n} + {b^n}}}{2}} \right) = \left( {\dfrac{{{8^{\dfrac{1}{3}}} + {{27}^{\dfrac{1}{3}}}}}{2}} \right)\]
\[ \Rightarrow \left( {\dfrac{{{a^n} + {b^n}}}{2}} \right) = \left( {\dfrac{{2 + 3}}{2}} \right)\]
On further simplification it becomes
\[ \Rightarrow \left( {\dfrac{{{a^n} + {b^n}}}{2}} \right) = 2.5 - - - (2)\]
Similarly, on putting the values in the right-hand side of the equation, we get
\[{\left( {\dfrac{{a + b}}{2}} \right)^n} = {\left( {\dfrac{{8 + 27}}{2}} \right)^{\dfrac{1}{3}}}\]
\[ \Rightarrow {\left( {\dfrac{{a + b}}{2}} \right)^n} = {\left( {17.5} \right)^{\dfrac{1}{3}}}\]
By using the calculator, we will find the value of the right-hand side of the above equation
\[ \Rightarrow {\left( {\dfrac{{a + b}}{2}} \right)^n} = 2.5962 - - - (3)\]
So, from equation \[(2)\] and equation \[(3)\] it is clear that equation \[(1)\] is true.

Note:
Here, one important point to note is that for equation \[(1)\] to be true, the value of \[n\] should be between zero and one i.e., \[0 \leqslant n \leqslant 1\] . Also note that for \[n = 0{\text{ and }}n = 1\] , that is for the upper and lower limits of \[n\], equality holds means that the left-hand side of equation \[(1)\] becomes equal to the right-hand side.