Q. 5

# Are there numbers x, y such that |x + y| < |x| + |y|?

Answer :

To prove : |x + y| < |x| + |y|

We know that, |x| and |y|

Therefore, 2|x||y|

Adding x^{2} + y^{2} to both sides,

We have, x^{2} + y^{2} + 2|x||y| x^{2} + y^{2} + 2xy

⇒ |x|^{2} + |y|^{2} + 2|x||y| x^{2} + y^{2} + 2xy

⇒ (|x| + |y|)^{2} (x + y)^{2}

⇒ (|x| + |y|)^{2} (|x + y|)^{2}

⇒ |x| + |y| |x + y|

We can also say that |x| + |y| > |x + y|

Therefore, this inequality holds true for all x and y.

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