The INEQUALITY quest!

Inequalities are important for olympiads and I.S.I., C.M.I. entrances. Sometimes they appear as standalone problems. However mostly they are clubbed with other problems (geometry, calculus etc). The Cheenta Inequality course is a 7-session-super-course to achieve excellence and efficiency in solving inequalities. The course work consists of 600 problems on inequalities with interdisciplinary treatment in certain special areas. For example while discussing triangular inequality or Euler's inequality we bring geometry in action. Similarly we discuss Lagrange's mean value theorem, idea of limit while discussing sequential inequalities, convexity of function and so on.
The 600 problems are sourced from:

• 70 problems from Little Mathematical Library
• 125 problems from Arthur Engel
• 100 problems from Excursion in Math and Challenges and Thrills of Precollege Math
• 300 problems from Inequalities: An Approach Through Problems By Venkatchala, International Math Olympiads and other national olympiads.

The course is resumed with certain trivial-looking ideas (natural numbers can be counted, ordered, square of real number is always non-negative, positive times positive is always positive). In the first stroke we cover inequalities related to Means, Bernoulli, Cauchy–Bunyakovsky–Schwarz, Holder. We also cover the simple cases of approximating sequential sums.
Then we move over to Jensen (convexity etc.) Rearrangement, Schur, Murihead and some other not-so-trivial-looking inequalities. Triangular inequality and application of calculus in understanding inequalities is discussed in the final phase.
The Kiran Kedlaya Inequality pack and similar courses available freely online were helpful. We tried to incorporate the best part of all of them. A course on inequality must be holistic as well as efficient. We must not loose our focus from problem-solving techniques. At the same-time the course must bear an elegance of the art-form.

IMO 1959 solution of Q4

Statement: Construct a right angled triangle with given hypotenuse c such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.


1st Solution:

Let |AB| = x and |AC| = x'

Then AD² = x.x'

Also x² + x'² = c² (since ABC is right angled at A).

Note that |AD| = c/2 since AD is the median of a right angled triangle with hypotenuse c units. (application of Apolonius Theorem)

Thus \(\frac{c^2}{4}=x\sqrt{c^2 - x^2}\) 

Hence we find 'x' in terms of c.

We draw a circle with diameter c. Cut off x = f(c) length from 'c' and construct the desired triangle.