Question

If \(x\) and \(y\) are positive real numbers satisfying \(x+y=102\), then the minimum possible value of \(2601\bigg(1+\frac{1}{x}\bigg)\bigg(1+\frac{1}{y}\bigg) \)is

Answer 1

\(AM≥GM≥HM\)

\(\frac{x+y}{2}≥\sqrt{xy}≥\frac{2}{\frac{1}{x}+\frac{1}{y}}\)

Given \(x+y=102\)

\(⇒ xy≤51^2 \;or\; \frac{1}{xy}≥\frac{1}{2601}\)

\(⇒\frac{ 1}{x}+\frac{1}{y}≥\frac{2}{51}\)

The minimum value of \(2601\bigg(1+\frac{1}{x}\bigg)\bigg(1+\frac{1}{y}\bigg)=(2601)\bigg(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\bigg)\)

= \(2601\bigg(1+\frac{2}{51}+\frac{1}{2601}\bigg)\)

= \(2704\)

Answer 2

Given,
\(x+y=102\)
As we know,
Arithmetic Mean = \(\frac{x+y}{2}\), Geometric Mean = \(\sqrt{xy}\) and AM≥GM
\(\frac{x+y}{2} \ge \sqrt{xy}\)

\(\frac{102}{2} \ge \sqrt{xy}\)
\(51 \ge \sqrt{xy}\)
\(51^2 \ge xy\)
\(2601 \ge xy\)

Now, we have to find maximum possible value of  \(2601\bigg(1+\frac{1}{x}\bigg)\bigg(1+\frac{1}{y}\bigg)\)
\(=2601(\frac{xy+y+x+1}{xy})\)

put the value of x+y and xy
\(=2601(\frac{2601+102+1}{2601})\)
\(=2704\)

So, the answer is 2704.