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121 - 130 of 315 for Compound InequalitiesSearch Results
Let A=a_(ik) be an arbitrary n×n nonsingular matrix with real elements and determinant |A|, then |A|^2<=product_(i=1)^n(sum_(k=1)^na_(ik)^2).
Given a positive sequence {a_n}, sqrt(sum_(j=-infty)^infty|sum_(n=-infty; n!=j)^infty(a_n)/(j-n)|^2)<=pisqrt(sum_(n=-infty)^infty|a_n|^2), (1) where the a_ns are real and ...
The hinge theorem says that if two triangles DeltaABC and DeltaA^'B^'C^' have congruent sides AB=A^'B^' and AC=A^'C^' and ∠A>∠A^', then BC>B^'C^'.
Let V be an inner product space and let x,y,z in V. Hlawka's inequality states that ||x+y||+||y+z||+||z+x||<=||x||+||y||+||z||+||x+y+z||, where the norm ||z|| denotes the ...
Given a convex plane region with area A and perimeter p, then |N-A|<p, where N is the number of enclosed lattice points.
For 0<=x<=pi/2, 2/pix<=sinx<=x.
A function f(x) satisfies the Lipschitz condition of order beta at x=0 if |f(h)-f(0)|<=B|h|^beta for all |h|<epsilon, where B and beta are independent of h, beta>0, and alpha ...
A function f is said to have a lower bound c if c<=f(x) for all x in its domain. The greatest lower bound is called the infimum.
If x takes only nonnegative values, then P(x>=a)<=(<x>)/a. (1) To prove the theorem, write <x> = int_0^inftyxP(x)dx (2) = int_0^axP(x)dx+int_a^inftyxP(x)dx. (3) Since P(x) is ...
For b>a>0, 1/b<(lnb-lna)/(b-a)<1/a.
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