Question

If $x,y,z$ are non-zero real number and $t=\frac{|x+y|}{|x|+|y|}+\frac{|x+z|}{|x|+|z|}+\frac{|y+z|}{|y|+|z|}$, then range of $t$ must be

• A. $0\le t\le 1$
• B. $0\le t\le 3$
• C. $1\le t\le 2$
• D. $1\le t\le 3$

Answer 1

The correct option is D $1\le t\le 3$

$\left|x\right|+\left|y\right|=|x+y|$ if $x$ and $y$ are of same sign, i.e. both $\ge 0$ or both $\le 0$

Therefore,

$\frac{|x+y|}{(\left|x\right|+\left|y\right|)}=1$ if $x$ and $y$ are of same sign.

From above, it follows that if $x,y$ and $z$ are all of same sign, then the equation is equals to $3$.

The other possibility is that two are of same sign and the third with the opposite sign.

Let $x$ and $y$ are of same sign with losing generality.

In that case,

$\frac{|x+y|}{(\left|x\right|+\left|y\right|)}=1$,

while $0\le \frac{|x+z|}{(\left|x\right|+\left|z\right|)}<1$ and

$0\le \frac{|y+z|}{(\left|y\right|+\left|z\right|)}<1$.

Thus established that:

(a) The upper limit is $3$ i.e. the value when $x,y$ and $z$ are all of same sign.

(b) The lower limit is at least $1$, since the same signed pair yield $1$, while any opposite signs pair yield at least $0$.

To establish $1$ as the lower limit, let $x=y=z$.

Then, the equation resolves to a value of $1$.

Hence, the answer is $1\le t\le 3$.