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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.

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