Answer
This question has a direct proof.
To prove: |x|+|y| $\geq$ |x+y|
If we take square of RHS we have,
$(|x|+|y|)^{2}$ = $|x|^{2}$ + $|y|^{2}$ + 2|x||y|......(1)
Considering the RHS of equation 1 we have,
($|x|^{2}$ + $|y|^{2}$ + 2|x||y|) $\geq$ $x^{2}$ + $y^{2}$ + 2xy (Because |x||y| $\geq$ xy)
Therefore from equation 1 we have,
$(|x|+|y|)^{2}$ $\geq$ $x^{2}$ + $y^{2}$ + 2xy
=> $(|x|+|y|)^{2}$ $\geq$ $(x+y)^{2}$
Since both sides are positive,
Therefore,
|x|+|y| $\geq$ |x+y|
Hence proved.
Work Step by Step
This question has a direct proof.
To prove: |x|+|y| $\geq$ |x+y|
If we take square of RHS we have,
$(|x|+|y|)^{2}$ = $|x|^{2}$ + $|y|^{2}$ + 2|x||y|......(1)
Considering the RHS of equation 1 we have,
($|x|^{2}$ + $|y|^{2}$ + 2|x||y|) $\geq$ $x^{2}$ + $y^{2}$ + 2xy (Because |x||y| $\geq$ xy)
Therefore from equation 1 we have,
$(|x|+|y|)^{2}$ $\geq$ $x^{2}$ + $y^{2}$ + 2xy
=> $(|x|+|y|)^{2}$ $\geq$ $(x+y)^{2}$
Since both sides are positive,
Therefore,
|x|+|y| $\geq$ |x+y|
Hence proved.
