Chapter 7.4, Problem 78HP

To determine

To find if the statement is true or false and justify the answer.

Answer to Problem 78HP

True

Explanation of Solution

Given:

  $x+y>\sqrt{x^2+y^2}$ when $x>0$ and $y>0$

Since x and y are both positive, each side of the inequality represents a positive number making the statement true.

To prove this,

Square the equation on both the sides.

  ${(x+y)}^2>{(\sqrt{x^2+y^2})}^2$

Simplifying the above equation,

  $x^2+y^2+2xy>x^2+y^2$

Subtract $x^2$ and $y^2$ on both sides.

  $2xy>0$

Conclusion:

Therefore, the given statement is true for all the values of x and y greater than zero.