EJC 2026 H3 Mathematics - Inequalities - Notes, Tutorial, and Solutions to Tutorial
Uploaded by noob12345 · 25 August 2026
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Text from the first pages1 H 3 M ATHEMATICS ( 9820 ) I NEQUALITIE S Note: H3 Inequalities is very different in nature from the Inequalities content in H2 Math. Example 1 Show that if a and b are real numbers such that 0 1a b , then 1a bb a+ + . Solution The Arithmetic Mean-Geometric Mean Inequality For any nonnegative real numbers 12, , , nx x x , 12 12 n n n x xn xx xx++ + , where the equality holds if and only if 12 nx x x= = = Proof In addendum. Example 2 (Do not use AM-GM) Prove that for ,0ab , 2 ab ab+ . Prove that for , , , 0a b c d , 4 4 a b c d abcd+ + + . Solution
2 Example 3 Find the minimum value of 2 361x x + for 0x . Solution Example 4 Show that 4bc b ca a d d+ + + where a, b, c, d are positive real numbers. Solution Example 5 Show that if x is a positive real number, 54 188 10xx − + . Solution
3 Example 6 Show that 2 2 2x y z xy yz zx + +++ where x, y and z are non-negative real numbers. Thought Process AM-GM tells us that 2 2 2 2 2 23 3 x y z zx y+ + . Does this help us? Solution Example 7 Show that 3 3 2 2 xx y y xy++ where x and y are non-negative real numbers. Show further that 11n n n n xxx y y y−− ++ for any positive integer n. Solution The Cauchy-Schwarz Inequality For any real numbers 12, , , nu u u and 12, , , nv v v , 2 22 1 1 1 n n n i i i i i i i vu v u = = = , where the equality holds if there exists a nonzero constant k such that iiu kv= for all 1, 2, ,in=
4 Proof In addendum. Example 8 Show that, if a, b and c are non-zero real numbers, ( ) 025 36 81 40abc ca b +++ + . Solution By Cauchy-Schwarz, Example 9 Find the maximum value of 23x y z++ , given that 2 2 2 1x y z+ + = . Thought Process Cauchy-Schwarz involves a product of two sums on the “large” side, and the square of a sum (of square-root terms) on the “small” side. It takes familiarity to be able to “guess” what goes where. Solution Example 10 Find the minimum value of 22ab+ , given that 3 4 15ab+= . Solution
5 Example 11 Let ,,x y z + . Prove that ( ) ( ) ( ) ( )3 3 3 2x x y y y z z z x x y z+ + + + + + + . Solution Example 12 Let , , ,a b c d + such that 1a b c d+ + + = . Find the minimum value of 1 1 1 1 bda c+ + + . Solution Example 13 Show that 2 2 2 2 x x y zy z z x x y yz + + +++ ++ for , , 0x y z . Thought Process How did Example 12 “get rid of” denominators? Solution
6 Example 14 Heron’s formula states that the area of a triangle with sides a, b and c is equal to ( )( )( )s s a s b s c− − − where 2s abc= ++ is the semiperimeter. A triangle with sides a, b and c has perimeter P and area T. (i) Show that 2 12 3 T P . (ii) Hence, show that ( ) 2 2 2 43abc T++ . Solution (i) (ii)
7 The Triangle Inequality Suppose the lengths of the sides of a triangle are a, b, c. Then the sum of the lengths of any two sides must be larger than the length of the third side, i.e. the following must all be true: a b c+ , b c a+ , c a b+ . This can be shown algebraically, but a powerful intuitive argument goes: the third side is the straight line segment joining its end points, and so must be the shortest path between its endpoints. In particular, the path that follows the other two sides is a longer path. A less common but equivalent way of stating this fact is that the length of any side must be longer than the difference of the lengths of the other two sides: a b c− , b a c− , c a b− . The inequality signs above are strict for non-degenerate triangles; when the triangle is degenerate in the sense that the three points lie on a straight line, the signs must be replaced with . Instead of considering points lying on some plane, we can consider points on a number line. Since these points form degenerate triangles, we can conclude that for any real numbers x and y, we must have x xyy+ + . Using induction, we can extend this to: For any real numbers 12, , , nx x x , 1 2 1 2 nnx x x x x x+++ ++ + , where the equality holds if and only if 12, , , nx x x are all non-negative Proof In addendum. Example 15 Show that for all , , .x y x y x y − − Hence, show that .x y x y− − Solution By the triangle equality, for all 12,xx , 1 2 1 2x x x x+ + For all ,xy , we define 1x xy=− and 2x y= Then x x y y−+ (*)x y x y− − (shown) Exchanging x and y in (*) , y x y x− − ( ) (**)x y x y− − − By (*) and (**), ( ) max ,x y x y x y x y− − − − = −
8 Example 16 Let a, b, c be the sides of a triangle. Prove that 2a b c b c c a a b+ + + + + . Solution Since a, b, c are the sides of a triangle, we have a b c+ , a c b+ and b c a+ (why are they strict inequalities?) Method 1 Method 2 Without loss of generality, let Keep in mind when attempting these tutorial questions: Our primary objective is to gain familiarity and mastery in applying AM -GM and Cauchy - Schwarz inequalities, as well as general inequality-solving and problem-solving techniques. Thus, resist to urge to look up solutions. At least 5 elegant proofs of Nesbitt’s Inequality (Q8) are just a few mouse clicks away, but looking them up works against our primary objective. The longer you spend thinking about a question, the more you gain out of it. Crucially, effort is seldom wasted: the experience gained from trying approaches that do not work is as important as finding what finally works – you build a sense of what “going astray” looks like! It is not unusual to spend days thinking about the harder problems here. Tutorial Questions 1 Find the minimum value of 249x x + where 0x . 2 Find the minimum value of 2 9 x x− where 9x . 3 Let a, b, c be positive real numbers such that 2019abc+ + = . State the maximum possible value of 3 12 3 12 3 12abc+ + + + + , and prove that this is a maximum. 4 Prove that 3 3 3 2 2 2x y z xz y y z x+ + + + for positive numbers ,,x y z . 5 Given that 1pqr+ + = , find the minimum value of 2 2 22p q r++ .
9 6 Harmonic Means. By considering the AM-GM inequality for positive real numbers 12, , , nb b b and the substitution 1 rarb = , show that 1 2 112 11 na a a n n na a a + ++ . 7 Let ,,abc + such that 1abc= . Find the minimum value of 1 1 1 1 1 1 ab bc ca a b c + + ++++ + + . 8 Nesbitt’s Inequality. Show that 3 2 a b c c a a b b c+ + + + + for positive real numbers a, b, c. 9 Titu’s Lemma. Let 12, , , na a a be real numbers and 12, , , nb b b be positive numbers. Show that ( ) 2222 2 1212 1 1 2 n n n n a aaaaa b b b b b b + + ++ + + + ++ . 10 Let , , 0abc . Prove that 3 3 3a b c abcbc ca ab+ + + + . 11 12, , , na a a are positive numbers such that 12 na a a s+ + + = . Show that 1 1 n i i i a n sa n= − − and ( ) 1 1 i i n i s na a n = − − . 12 Find the minimum value of 22 36 36x y y x++ + where ,xy . 13 ,,x y z are positive real numbers. Show that 3 23x y x yz xyz + , and hence show that 3 x x y z y xyz yz zx +++ + . 14 Let a, b, c be nonnegative real numbers such that 2 2 2 1 1 1 21 1 1a bc+ + =+ + + . State the value of 2 2 2 2 2 21 1 1 abc a bc+++ + + and hence show that 3 2ab bc ca++ . 15 Let a, b, c be positive numbers such that 2 2 2 3abc+ + = . (i) Show that 4 4 4 3abc++ . (ii) Deduce from (i) that 2 2 2 2 2 2 3a b b c c a+ + . (iii) Hence
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