Given $5n$ real numbers $r_i, s_i, t_i, u_i, v_i \geq 1 (1 \leq i \leq n)$, let $R = \frac {1}{n} \sum_{i=1}^{n} r_i$, $S = \frac {1}{n} \sum_{i=1}^{n} s_i$, $T = \frac {1}{n} \sum_{i=1}^{n} t_i$, $U = \frac {1}{n} \sum_{i=1}^{n} u_i$, $V = \frac {1}{n} \sum_{i=1}^{n} v_i$. Prove that $\prod_{i=1}^{n}\frac {r_i s_i t_i u_i v_i + 1}{r_i s_i t_i u_i v_i - 1} \geq \left(\frac {RSTUV +1}{RSTUV - 1}\right)^n$.
Problem
Source: China TST 1994, problem 4
Tags: inequalities, logarithms, function, algebra, domain, linear algebra, matrix
keira_khtn
20.05.2005 09:31
Applying jensen's ineq for five multivariables!!!
Singular
22.05.2005 04:23
can you please explain the multivariable jensen? i cant find it anywhere
Fedor Petrov
22.05.2005 09:59
Let $X=(x_1,x_2,x_3,x_4,x_5)$, $F(X)=\ln \frac{x_1x_2x_3x_4x_5+1}{x_1x_2x_3x_4x_5-1}$. Then $F(x)$ is a convex function in a domain $(0,+\infty)^5$. For proving this, we must assure that Hessian $H(X)=(\partial^2 F/\partial x_i\partial x_j)_{i,j=1}^5$ is non-negative defined in every point $(x_1,x_2,x_3,x_4,x_5)$. We have $\partial^2 F/\partial x_i\partial x_j=\frac{2p(p^2+1)}{x_ix_j(p^2-1)^2}, i\ne j$ and $\partial^2 F/\partial x_i^2=\frac{4p^3}{x_i^2(p^2-1)^2}.$ (here $p=x_1x_2x_3x_4x_5$) So, we have to prove that the matrix with diagonal entries $2p^2$ and not-diagonal elements $p^2+1$ is not-negative defined (we multipled rows and columns by corresponding $x_i$ and then multiply by some constant). But it is clear: matrix with equal entries is non-negative defined, and our matrix differs from it by a positive scalar matrix. So, just apply Jensen for this function.
who
19.06.2006 13:33
,holder does it nicely.
soryn
29.09.2024 01:48
Is possible a elementary solution?
EthanWYX2009
29.09.2024 04:12
Of course Jensen kills the problem immediately, here is a fun solution:
soryn
29.09.2024 06:34
Very nice, thanks!