retagged by
507 views
2 votes
2 votes
If $\textsf{x}$ and $\textsf{y}$ are positive real numbers satisfying $\textsf{x+y = 102},$ then the minimum possible value of $\textsf{2601} \left( 1 + \frac{1}{\textsf{x}} \right) \left( 1 + \frac{1}{\textsf{y}} \right)$ is
retagged by

1 Answer

1 votes
1 votes
Given that, $x$ and $y$ are positive real numbers, satisfying $x+y = 102.$

We know that, $\boxed{\text{AM} \geq \text{GM} \geq \text{HM}}$

$ \Rightarrow \frac{x+y}{2} \geq \sqrt{xy} \geq \dfrac{2}{\frac{1}{x} + \frac{1}{y}} $

$ \Rightarrow \frac{102}{2} \geq \sqrt{xy} \geq \dfrac{2}{\frac{1}{x} + \frac{1}{y}} \quad \longrightarrow (1)$

Take the first two terms, from equation $(1).$

$ 51 \geq \sqrt{xy} $

$ \Rightarrow xy \leq 51^{2} $

$ \Rightarrow \boxed{xy \leq 2601} $

Taking the last two terms, from equation $(1).$

$ \sqrt{xy} \geq \dfrac{2}{\frac{1}{x} + \frac{1}{y}} $

$ \Rightarrow 51 \geq \dfrac{2}{\frac{1}{x} + \frac{1}{y}} $

$ \Rightarrow \boxed{\frac{1}{x} + \frac{1}{y} \geq \frac{2}{51}} $

Now, the minimum possible value of $ 2601 \left( 1 + \frac{1}{x} \right) \left( 1 + \frac{1}{y} \right) = 2601 \left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \right) $

$\qquad  = 2601 \left( 1 + \frac{2}{51} + \frac{1}{2601} \right)  = 2601 \left( \frac{2601 + 102 + 1}{2601} \right)  = 2704 $

$\therefore$ The minimum possible value is $2704.$

Correct Answer$: 2704$
edited by
Answer:

Related questions

2 votes
2 votes
1 answer
1
soujanyareddy13 asked Sep 17, 2021
581 views
For real $\textsf{x}$ , the maximum possible value of $ \frac{x}{\sqrt{1+x^{4}}}$ is $ \frac{1}{\sqrt{3}}$$1$$\frac{1}{\sqrt{2}}$$\frac{1}{2}$
3 votes
3 votes
1 answer
2
soujanyareddy13 asked Sep 17, 2021
560 views
The number of integers that satisfy the equality $\left( x^{2} – 5x + 7 \right)^{x+1} = 1$ is $2$$3$$5$$4$
2 votes
2 votes
1 answer
5
soujanyareddy13 asked Sep 17, 2021
649 views
How many $4$-digit numbers, each greater than $1000$ and each having all four digits distinct, are there with $7$ coming before $3.$