If x, y, z are positive integers, then value of expression (x + y)(y + z)(z + x) is

Question

If x, y, z are positive integers, then value of expression (x + y)(y + z)(z + x) is (A) ≠ 8xyz (B)  ge 8xyz (C)  le 8 xyz (D) 4xyz

Tardigrade Admin · Accepted Answer

We know that, A.M. ge G.M. ⇒ (x+y/2) ge √xy, (y+z/2) ge √yz and (z+x/2) ge√zx Multiplying the three inequalities, we get (x+y/2)⋅ (y+z/2)⋅( x+z/2) ge√ (xy)(yz)(zx) or, (x+y)(y+z)(z+x) ge 8xyz

Q. If $x, y, z$ are positive integers, then value of expression $(x + y) (y + z) (z + x)$ is

$\neq = 8 x yz$

$\geq 8 x yz$

$\leq 8 x yz$

$4 x yz$

We know that, $A . M . \geq G . M$ .
$\Rightarrow \frac{x + y}{2} \geq x y, \frac{y + z}{2} \geq yz$ and
$\frac{z + x}{2} \geq z x$
Multiplying the three inequalities, we get
$\frac{x + y}{2} \cdot \frac{y + z}{2} \cdot \frac{x + z}{2} \geq (x y) (yz) (z x)$
or, $(x + y) (y + z) (z + x) \geq 8 x yz$

Q. If x,y,z are positive integers, then value of expression (x+y)(y+z)(z+x) is

=8xyz

≥8xyz

≤8xyz

4xyz

We know that, A.M.≥G.M. ⇒2x+y​≥xy​,2y+z​≥yz​ and 2z+x​≥zx​ Multiplying the three inequalities, we get 2x+y​⋅2y+z​⋅2x+z​≥(xy)(yz)(zx)​ or, (x+y)(y+z)(z+x)≥8xyz

Q. If $x, y, z$ are positive integers, then value of expression $(x + y) (y + z) (z + x)$ is

$\neq = 8 x yz$

$\geq 8 x yz$

$\leq 8 x yz$

$4 x yz$