Inequalities path · I12

Before this lesson: Arithmetic, geometric and harmonic means, Triangle inequalities

Your goal: Check both the algebraic bound and the geometric attainability.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Inequalities in geometry: the key idea

Geometry adds feasibility conditions to algebra. Three positive lengths form a nondegenerate triangle exactly when each is less than the sum of the other two. With semiperimeter s=(a+b+c)/2, the variables x=s−a,y=s−b,z=s−c are positive and satisfy a=y+z,b=z+x,c=x+y. This substitution builds the triangle inequalities into an algebraic problem. A claimed equality configuration must exist geometrically. Bounds from AM-GM, Cauchy and area formulas become useful after translating the diagram into lengths or ratios.

A worked example

Can sides 2,3,5 form a triangle?

No nondegenerate triangle exists: 2+3=5 gives a straight, collapsed configuration. Triangle inequalities must be strict.

Your turn: change one thing

For a triangle of perimeter 12, prove ab+bc+ca≤48.

Try this on paper before opening the explanation.

Compare your reasoning

Since a+b+c=12, (a+b+c)²≥3(ab+bc+ca), giving at most 48. Equality is a=b=c=4, which is a valid equilateral triangle.

Pause and check

A trap to avoid: Accepting equality at an impossible or degenerate triangle.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Prove that the perimeter of a nondegenerate triangle exceeds twice any one side.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

The triangle inequality b+c>a gives a+b+c>2a. Repeat for either other side. Strictness follows from nondegeneracy; a collinear limiting configuration is not an equality triangle.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. Which lengths form a nondegenerate triangle?

Foundation

  1. 1,2,3
  2. 2,2,5
  3. 1,1,2
  4. 3,4,5
Hint

Check the sum of the smaller two.

Answer and reasoning

3,4,5. 3+4>5 and the other two inequalities also hold.

2. For a valid triangle, s−a has which sign?

Foundation

  1. Any sign
  2. Positive
  3. Zero
  4. Negative
Hint

Rewrite it as (b+c−a)/2.

Answer and reasoning

Positive. The triangle inequality makes the numerator positive.

3. A triangle has perimeter 18. Every side is strictly below what number?

Core

  1. 18
  2. 3
  3. 9
  4. 6
Hint

Perimeter exceeds twice each side.

Answer and reasoning

9. 2a<18 gives a<9.

4. For triangle perimeter 12, maximum ab+bc+ca?

Core

  1. 48
  2. 36
  3. 72
  4. 144
Hint

Use the square of the perimeter.

Answer and reasoning

48. 144≥3(ab+bc+ca); equality at sides 4,4,4 gives 48.

5. Under x=s−a,y=s−b,z=s−c, a equals what?

Stretch

  1. y+z
  2. x+y+z
  3. x−y
  4. 2x
Hint

Add the definitions of y and z.

Answer and reasoning

y+z. y+z=2s−b−c=a.

6. Why must equality be checked on the diagram too?

Stretch

  1. Algebra never applies to geometry
  2. All triangles are equal
  3. A sketch proves feasibility
  4. An algebraic equality case may violate geometric constraints
Hint

Lengths must describe an allowed configuration.

Answer and reasoning

An algebraic equality case may violate geometric constraints. Positivity, triangle inequalities and incidence conditions must all be satisfied.

7. For a triangle, the semiperimeter s equals what?

Foundation

  1. a/2
  2. (a+b+c)/2
  3. a+b+c
  4. abc
Hint

It is half the perimeter.

Answer and reasoning

(a+b+c)/2. The sum of the three side lengths is divided by two.

8. A triangle has sides 4,5,6. What is s−6?

Core

  1. 1/2
  2. 9/2
  3. 3/2
  4. 3
Hint

First find s=15/2.

Answer and reasoning

3/2. 15/2−6=3/2.

9. When does ab+bc+ca reach its largest value for a fixed positive perimeter?

Stretch

  1. At an equilateral triangle
  2. At a collapsed triangle
  3. At any right triangle
  4. At the longest possible side
Hint

Use the sum-of-squares equality case.

Answer and reasoning

At an equilateral triangle. The bound from a²+b²+c²≥ab+bc+ca is attained exactly when a=b=c.

Choose your next step

Continue to Jensen’s inequality. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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