Home » Uncategorized » Inequality 24(Moldavian TST 2002)

Inequality 24(Moldavian TST 2002)

Problem:

Positive real numbers \displaystyle a,b,x_1,x_2,...,x_n satisfy the condition \displaystyle x_1+x_2+...+x_n=1. Prove that

\displaystyle \frac{x^{3}_{1}}{ax_1+bx_2}+\frac{x^{3}_{2}}{ax_2+bx_3}+...+\frac{x^{3}_{n}}{ax_n+bx_1}\geq \frac{1}{n(a+b)}.

Solution:

From Holder’s inequality we have that:

\displaystyle (1+1+...+1)\left(\sum^{n}_{i=1}\frac{x^{3}_{1}}{ax_i+bx_{i+1}}\right)\left[\sum^{n}_{i=1}\left(ax_i+bx_{i+1}\right)\right]\geq \left(\sum^{n}_{i=1}x_i\right)^{3}=1.

So, it remains to prove

\displaystyle \sum^{n}_{i=1}\frac{x^{3}_{i}}{ax_i+bx_{i+1}}\geq \frac{1}{n\sum^{n}_{i=1}ax_i+bx_{i+1}}\geq \frac{1}{n(a+b)}.

But \displaystyle \sum^{n}_{i=1}ax_i+bx_{i+1}=a+b, Q.E.D.

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