Question

Assertion :STATEMENT -1 : If $x,y,z$ are the sides of a triangle such that $x+y+z=1$, then $\left[\frac{2x-1+2y-1+2z-1}{3}\right]\ge \left(\right(2x-1\left)\right(2y-1\left)\right(2z-1){)}^{1/3}$ Reason: STATEMENT-2 : For positive numbers their A.M., G.M. and H.M. satisfy the relation $A.M.>G.M.>H.M.$

• A. Statement -1 is True, Statement -2 is True ; Statement -2 is a correct explanation for Statement -1
• B. Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1
• C. Statement -1 is True, Statement -2 is False
• D. Statement -1 is False, Statement -2 is True

Answer 1

The correct option is D Statement -1 is False, Statement -2 is True

$x+y>z$ i.e. $1-z>z$ i.e. $1>2z$
Similarly, $1>2x$ and $1>2y$
$\therefore 2x-1<0,2y-1<0,2z-1<0$
We know that for positive numbers A.M $\ge$G.M, then
$\frac{(1-2x)+(1-2y)+(1-2z)}{3}\ge {\left\{\right(1-2x\left)\right(1-2y\left)\right(1-2z\left)\right\}}^{1/3}$
Now, multiplying both sides by $-1$, we get
$\frac{(1-2x)+(1-2y)+(1-2z)}{3}\le {\left\{\right(2x-1\left)\right(2y-1\left)\right(2z-1\left)\right\}}^{1/3}$
$\therefore$ statement-1 is false $\therefore$ statement-2 is true
Hence, option 'D' is correct.