State True or False.
If x, y are positive real numbers such that x + y = 1, then $(1 + \frac{1}{x}) (1 + \frac{1}{y}) ⩾ 9$ .

True

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False

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Solution

The correct option is A True
For positive real x,y applying $A M \geq G M$
$\frac{x + y}{2} \geq \sqrt{x y}$
Since $x + y = 1$ ,
we have $\frac{1}{2} \geq \sqrt{x y}$
$\Rightarrow$ $1 \geq 2 \sqrt{x y}$
$1 \geq 4 x y$
$2 \geq 8 x y$
$1 + 1 \geq 8 x y$
(i.e.,) $(x + y + 1) \geq 8 x y$
$x + y + x y + 1 \geq 9 x y$
$(x + 1) (y + 1) \geq 9 x y$
$\frac{(x + 1) (y + 1)}{x y} \geq 9$
$\Rightarrow$ $(1 + \frac{1}{x}) (1 + \frac{1}{y}) ⩾ 9$

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