For any two real numbers $x & y,$ the inequality that holds true is

| x + y | \leq | x | - | y |

No worries! We‘ve got your back. Try BYJU‘S free classes today!

| x + y | \leq | x | + | y |

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

| x + y | \geq | x | + | y |

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $| x + y | \geq | x | + | y |$
Given two real numbers $x & y .$
To find the relationship between $| x |, | y | & | x + y |$
Let's consider the case where $x, y \geq 0$
$\Rightarrow | x | = x & | y | = y$
Also, $| x + y | = x + y = | x | + | y |$
$\Rightarrow | x + y | = | x | + | y | \forall x, y \geq 0$

Let's consider the case where $x \geq 0 & y < 0$
$\Rightarrow | x | = x & | y | = - y$
$\Rightarrow | x | + | y | = x + (- y)$
Whose value is greater than $| x + y |$
$\Rightarrow | x + y | < | x | + | y | \forall x \geq 0, y < 0$

Let's consider the case where $x, y \leq 0$
$\Rightarrow | x | = - x & | y | = - y$
$\Rightarrow | x | + | y | = - x + (- y)$
Also, $| x + y | = - (x + y) = | x | + | y |$
$\Rightarrow | x + y | = | x | + | y | \forall x, y \leq 0$
Thus for any real number $x, y$
$| x + y | \leq | x | + | y |$

Suggest Corrections

Similar questions

x,

\frac{- x}{4} - 3 < - x

10 + y \geq 10

x - 10 \geq - 10

x

y

x, y

x y \leq 0,

| x - y | = | x | + | y | .

The set of real values of $x$ for which inequality
${log}_{2} (x^{2} - x - 6) + {log}_{0.5} (x - 3) < 2 {log}_{2} 3$
holds true is

x, y & y \neq 0

∣ ∣ ∣ \frac{x}{y} ∣ ∣ ∣ = \frac{| x |}{| y |} .

Join BYJU'S Learning Program