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Tag Archives: Inequality
Fix . Apply the matrix on thus:
Since is exactly the – th column of the previous equality can be rerwritten as
Since this holds for all we get and the left inequality follows.
For a random unit vector the coordinate of the vector is . It follows from Cauchy – Schwartz that
Summing over all ‘s till we find we conclude that, for every unit vector , it holds that is less than the right hand side. Taking supremum with respect to all the right hand side inequality follows.
Let . Prove that:
Due to homogeneity we may assume . Thus there exist positive such that
Let be a triangle. Prove that
The function is convex, thus:
The result follows. In fact, something a bit stronger holds. Let denote the inradius , the circumradius and the semiperimeter. Then,
in view of the known identities
Show that in any triangle with area the following holds:
Let is the semiperimeter , the circumradius and the inradius. From the law of sines we find
as well as
Substitute the preceding equalities into the last inequality and simplifying we obtain
the last inequality becomes
which is true by the rearrangement inequality.