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Bennett's inequality, an upper bound on the probability that the sum of independent random variables deviates from its expected value by more than any specified amount Bhatia–Davis inequality , an upper bound on the variance of any bounded probability distribution
For instance, to solve the inequality 4x < 2x + 1 ≤ 3x + 2, it is not possible to isolate x in any one part of the inequality through addition or subtraction. Instead, the inequalities must be solved independently, yielding x < 1 / 2 and x ≥ −1 respectively, which can be combined into the final solution −1 ≤ x < 1 / 2 .
Since all the inequalities are in the same form (all less-than or all greater-than), we can examine the coefficient signs for each variable. Eliminating x would yield 2*2 = 4 inequalities on the remaining variables, and so would eliminating y. Eliminating z would yield only 3*1 = 3 inequalities so we use that instead.
Many important inequalities can be proved by the rearrangement inequality, such as the arithmetic mean – geometric mean inequality, the Cauchy–Schwarz inequality, and Chebyshev's sum inequality. As a simple example, consider real numbers : By applying with := for all =, …,, it follows that + + + + + + for every permutation of , …,.
In general, the variational inequality problem can be formulated on any finite – or infinite-dimensional Banach space. The three obvious steps in the study of the problem are the following ones: Prove the existence of a solution: this step implies the mathematical correctness of the problem, showing that there is at least a solution.
The inequality was first proven by Grönwall in 1919 (the integral form below with α and β being constants). [1] Richard Bellman proved a slightly more general integral form in 1943. [2] A nonlinear generalization of the Grönwall–Bellman inequality is known as Bihari–LaSalle inequality. Other variants and generalizations can be found in ...
Proof [2]. Since + =, =. A graph = on the -plane is thus also a graph =. From sketching a visual representation of the integrals of the area between this curve and the axes, and the area in the rectangle bounded by the lines =, =, =, =, and the fact that is always increasing for increasing and vice versa, we can see that upper bounds the area of the rectangle below the curve (with equality ...
Consider the sum = = = (). The two sequences are non-increasing, therefore a j − a k and b j − b k have the same sign for any j, k.Hence S ≥ 0.. Opening the brackets, we deduce: