5.2.1.2. Maximum value, minimum value.
Exercise 5.7. Find the minimum value of the function:
Considering the domain .
Solution:
Apply the consequence of Holder's octal equality to two sequences of numbers ,
and
We get
On the other hand:
From (1) and (2) it follows that:
Applying the consequence of Holder's octave again to the two sequences x, y, z, t and y, z, t, x we get:
From (3) and (4) it follows that:
Exercise 5.7. Given tetrahedron ABCD. P is an arbitrary point in the tetrahedron. Let , be the projection of P onto the faces BCD,ACD,ABD and ABC. Let S and r be the total area and radius of the sphere inscribed in the tetrahedron, respectively. Prove:
Solution:
As a result of the Holder inequality, we have:
(T is the left hand side of the octal of equality to be proved)
On the other hand:
Exercise 5.8 Solve the equation:
Solution:
Applying the consequence of Holder's octave, we have:
The "=" sign occurs if and only if:
So the solution of the given equation is .
5.2.4. Applications in arithmetic.
Exercise 5.9. Let be real numbers satisfying: .Prove that:
Solution:
Consider two sequences and .
As a consequence of Hölder inequality, we have:
Or
∀k>1, we have:
Adding each side of equality (2) from k=2 to k=n+1, we get:
From (1), (3) deduce:
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