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Drawing Inequalities

Suppose we have an inequality x y. The line represented by the equation x = y is given as follows:

The points (1,1), (2,2), (3,3) lie on it. But (1,2), (-1,-2), (3,2) does not lie on it.

Substitute any point in the equation, say, (1,2). Does it satisfy the inequality? Since 1<2, it does. Hence the point (1, 2) will lie in the shaded portion of the line.


  This means that all points satisfying the inequality must lie in the shaded portion of the graph.

All points not fulfilling the inequality, for instance, (2,1) does not fulfil x y, will lie in the portion of the line which is not shaded.

We can thus draw the graph of any inequality.


Illustration 9: Which inequality is represented by the following graph:



Solution: The line represented above has the equality: 3y - 2x = 6.

Take any point, say (0,6). Substitute in the equation, we get: 18>6. Hence the shaded portion will be represented by the inequality 3y - 2x 6.


Elementary properties of inequalities:

i) If a > b and b > c, then a > c.

ii) If a > b, then a + m > b + m, for any real number m.

iii) If a > b, then am > bm for m > 0 and am < bm for m < 0.

that is, when we multiply both sides of the inequality by a negative quantity, the sign of the inequality reverses.

iv) If a 0 and b 0 and a > b, then (1/b) < (1/b).

v) (a + c + e + ……) /(b + d + f +…..) is greater than the least and less than the greatest of the fractions a/b, c/d, e/f, …

vi) If a > x, b > y, c > z, …. Then a + b + c +… > x + y + z + … and abc …> xyz …; abc and xyz should be all non negative numbers.

vii) a² + b² + c² ab + bc + ca

viii) (n!)² nn

ix) For any positive integer n, 2 (1 + 1/n)n â‰⁄ 3

x) a²b + b²c + c²a 3abc

xi) a/b + b/c + c/d + d/a > 4

xii) If x > 0 and a > b > 0; then ax > bx


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