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Algebra Difficulty 6.5 National olympiad Prove it

Theorem 13.11 Let u1,u2,,un,tu_{1}, u_{2}, \cdots, u_{n}, t be non-negative real numbers, 2m,nN2 \leqslant m, n \in \mathbf{N}, and let the polynomial f(y)f(y) in yy be defined as
f(y)=j1,j2,,jn1Y0,1,,m1(y[u1m+ζm(j1)u2m++ζmjn1munm]m)f(y)=\prod_{j_{1}, j_{2}, \cdots, j_{n-1}} \prod_{Y_{0,1, \cdots, m-1}}\left(y-\left[\sqrt[m]{u_{1}}+\zeta_{m}^{\left(j_{1}\right)} \sqrt[m]{u_{2}}+\cdots+\zeta_{m}^{j_{n-1} m} \sqrt[m]{u_{n}}\right]^{m}\right)

Then the inequality
u1m+u2m++unmtm\sqrt[m]{u_{1}}+\sqrt[m]{u_{2}}+\cdots+\sqrt[m]{u_{n}} \leqslant \sqrt[m]{t}

holds if and only if the following r=mn1r=m^{n-1} real numbers c0,c1,,cr1c_{0}, c_{1}, \cdots, c_{r-1} satisfy
c0=f(t)0,c1=f(1)(t)0,,cr1=f(r1)(t)0c_{0}=f(t) \geqslant 0, c_{1}=f^{(1)}(t) \geqslant 0, \cdots, c_{r-1}=f^{(r-1)}(t) \geqslant 0

where f(k)f^{(k)} is the kk-th derivative of ff, k=1,2,,r1k=1,2, \cdots, r-1.

Solution

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.