INEQUALITY'S BLOG

July 26, 2010

Van Khea’s inequality

Filed under: the new math — KKKVVV @ 5:55 pm

Let a, b, x, y>0, \alpha\geq 1, \beta, \gamma \geq 0 ; \alpha-\beta-\gamma=1. Prove that:

\displaystyle \frac{a^{\alpha}}{x^{\beta}p^{\gamma}}+\frac{b^{\alpha}}{y^{\beta}q^{\gamma}}\geq \frac{(a+b)^{\alpha}}{(x+y)^{\beta}(p+q)^{\gamma}}

Let x_i, y_i, z_i>0 ; i=1, 2, ..., n and \alpha\geq 1, \beta, \gamma \geq 0 such that \alpha -\beta -\gamma=1. We get:

\displaystyle \frac{x_1^{\alpha}}{y_1^{\beta}z_1^{\gamma}}+\frac{x_2^{\alpha}}{y_2^{\beta}z_2^{\gamma}}+...+\frac{x_n^{\alpha}}{y_n^{\beta}z_n^{\gamma}}\geq \frac{(x_1+x_2+...+x_n)^{\alpha}}{(y_1+y_2+...+y_n)^{\beta}(z_1+z_2+...+z_n)^{\gamma}}

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