Question

If x and y belong to real numbers, show that
|x+y|/1+|x+y|<=|x|/1+|x|+|y|/1+|y|

Answer 1

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QUESTION

If $x$ and $y$ belong to real numbers, show that

$$\frac{|x + y|}{1 + |x + y|} \leq \frac{|x|}{1 + |x|} + \frac{|y|}{1 + |y|}$$

SOLUTION

Firstly it must be said that the

$$\forall x, y \in \Re \quad |x| \geq 0, |y| \geq 0 \Longrightarrow 1 + |x| \geq 1, 1 + |y| \geq 1, 1 + |x + y| \geq 1$$

The proof will be constructed as follows: make a chain of identical transformations to get the performance of a simple inequality which data $x, y \in \Re$ obvious. As the $|x| + 1 \geq 1$ counter is positive, multiplying both sides of the inequality in the $|x| + 1$ will not change the sign of inequality

$$\begin{array}{l} \frac{|x + y|}{1 + |x + y|} \leq \frac{|x|}{1 + |x|} + \frac{|y|}{1 + |y|} \left| (1 + |x|) * (1 + |y|) * (1 + |x + y|) \right. \Longleftrightarrow \\ \frac{|x + y| * (1 + |x|) * (1 + |y|) * (1 + |x + y|)}{1 + |x + y|} \leq \\ \leq \frac{|x| * (1 + |x|) * (1 + |y|) * (1 + |x + y|)}{1 + |x|} + \frac{|y| * (1 + |x|) * (1 + |y|) * (1 + |x + y|)}{1 + |y|}; \\ \end{array}$$ $$\begin{array}{l} (1 + |x|) * (1 + |y|) * |x + y| \leq |x| * (1 + |y|) * (1 + |x + y|) + |y| * (1 + |x|) * (1 + |x + y|); \\ (1 + |x| + |y| + |x| * |y|) * |x + y| \leq |x| * (1 + |y| + |x + y| + |y| * |x + y|) + |y| * (1 + |x| + |x + y| + |x| * |x + y|); \\ |x + y| + |x| * |x + y| + |y| * |x + y| + |x| * |y| * |x + y| \leq |x| + |x| * |y| + |x| * |x + y| + |x| * |y| * |x + y| + \\ + |y| + |y| * |x| + |y| * |x + y| + |y| * |x| * |x + y|; \\ |x + y| \leq |x| + |y| + |x| * |y| * |x + y| + 2 * |x| * |y| \end{array}$$

Using the triangle inequality

$$|x + y| \leq |x| + |y|, \quad \forall x, y \in \Re$$ $$|x + y| \leq |x| + |y| + |x| * |y| * |x + y| + 2 * |x| * |y| \iff 0 \leq |x| * |y| * |x + y| + 2 * |x| * |y|$$

As the last inequality $0 \leq |x| * |y| * |x + y| + 2 * |x| * |y| \ \forall x, y \in \Re$ is satisfied, then the initial and inequality

$$\frac{|x + y|}{1 + |x + y|} \leq \frac{|x|}{1 + |x|} + \frac{|y|}{1 + |y|} \quad \forall x, y \in \Re$$