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# Inequality 52 (Unknown Author)

Problem:

Let $\displaystyle x_1,x_2,...,x_n$ be positive real numbers with sum equal to $\displaystyle 1$. Find the minimum value of the expression

$\displaystyle A=\max\left\{\frac{x_1}{1+x_1},\frac{x_2}{1+x_1+x_2},...,\frac{x_n}{1+x_1+x_2+...+x_n}\right\}$.

Solution:

From the definition of $\displaystyle A$, we have that

$\displaystyle A\geq \frac{x_1}{1+x_1}, A\geq \frac{x_2}{1+x_1+x_2},...,A\geq \frac{x_n}{1+x_1+x_2+...+x_n}$.

Solving the 1st Inequality we get that

$\displaystyle x_1\leq \frac{A}{1-A}$.

Solving now the 2nd Inequality we have that

$\displaystyle A\geq \frac{x_2}{1+x_1+x_2}\geq \frac{x_2}{1+\frac{A}{1-A}+x_2}=\frac{x_2}{\frac{1}{1-A}+x_2}$.

The last relation reduces to the

$\displaystyle x_2\leq \frac{A}{(1-A)^2}$.

So, we see now that $\displaystyle x_k\leq \frac{A}{(1-A)^k}$ for $\displaystyle k=1,2,...,n$.

Summing up these $\displaystyle n$ Inequalities we acquire

$\displaystyle x_1+x_2+...+x_n\leq \frac{A}{1-A}+\frac{A}{(1-A)^2}+...+\frac{A}{(1-A)^n}$

or, due to the hypothesis we now have that

$\displaystyle \frac{A}{1-A}+\frac{A}{(1-A)^2}+...+\frac{A}{(1-A)^n}\geq 1$.

If we solve this last Inequality it gives us the result $\displaystyle A\geq 1-\frac{1}{\sqrt[n]{2}}$, which is also and the value we are searching, Q.E.D.