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Difference between revisions of "Geometric inequality"

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Proof: <math>\sin</math> is a [[concave function]] from <math>0\le \theta \le \pi</math>. Therefore we may use [[Jensen's inequality]]: <math>\frac{\sin{A}+\sin{B}+\sin{C}}{3}\le \sin{\left(\frac{A+B+C}{3}\right)}=\frac{\sqrt{3}}{2}</math>
 
Proof: <math>\sin</math> is a [[concave function]] from <math>0\le \theta \le \pi</math>. Therefore we may use [[Jensen's inequality]]: <math>\frac{\sin{A}+\sin{B}+\sin{C}}{3}\le \sin{\left(\frac{A+B+C}{3}\right)}=\frac{\sqrt{3}}{2}</math>
  
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Alternatively, we may use a method that can be called "perturbation". If we let all the angles <math>A,B,C</math> be equal, we prove that if we make one angle greater and the other one smaller, we will decrease the total value of the expression. To prove this, all we need to show is if <math>0<A,B<180</math>, then <math>\sin(A+B)+\sin(A-B)<2\sin A</math>. This inequality reduces to <math>2\sin A \cos B<2\sin A</math>, which is equivalent to <math>cos B<1</math>. Since this is always true for <math>0<B<180</math>, this inequality is true. Therefore, the maximum value of this expression is when <math>A=B=C=60</math>, which gives us the value <math>\sin{A}+sin {B}+sin {C}=\frac{3\sqrt{3}}{2}</math>.
+
Alternatively, we may use a method that can be called "perturbation". If we let all the angles <math>A,B,C</math> be equal, we prove that if we make one angle greater and the other one smaller, we will decrease the total value of the expression. To prove this, all we need to show is if <math>0<A,B<180</math>, then <math>\sin(A+B)+\sin(A-B)<2\sin A</math>. This inequality reduces to <math>2\sin A \cos B<2\sin A</math>, which is equivalent to <math>\cos B<1</math>. Since this is always true for <math>0<B<180</math>, this inequality is true. Therefore, the maximum value of this expression is when <math>A=B=C=60</math>, which gives us the value <math>\sin{A}+\sin{B}+\sin {C}=\frac{3\sqrt{3}}{2}</math>.
  
 
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Revision as of 16:05, 8 February 2014

A geometric inequality is an inequality involving various measures (angles, lengths, areas, etc.) in geometry.

Pythagorean Inequality

The Pythagorean Inequality is a generalization of the Pythagorean Theorem. The Theorem states that in a right triangle with sides of length $a \leq b \leq c$ we have $a^2 + b^2 = c^2$. The Inequality extends this to obtuse and acute triangles. The inequality says:

For an acute triangle with sides of length $a \leq b \leq c$, $a^2+b^2>c^2$. For an obtuse triangle with sides $a \leq b \leq c$, $a^2+b^2<c^2$.

This inequality is a direct result of the Law of Cosines, although it is also possible to prove without using trigonometry.

Triangle Inequality

The Triangle Inequality says that the sum of the lengths of any two sides of a nondegenerate triangle is greater than the length of the third side. This inequality is particularly useful and shows up frequently on Intermediate level geometry problems. It also provides the basis for the definition of a metric space in analysis.

Isoperimetric Inequality

The Isoperimetric Inequality states that if a figure in the plane has area $A$ and perimeter $P$, then $\frac{4\pi A}{P^2} \le 1$. This means that given a perimeter $P$ for a plane figure, the circle has the largest area. Conversely, of all plane figures with area $A$, the circle has the least perimeter.

Trigonometric Inequalities

  • In $\triangle ABC$, $\sin{A}+\sin{B}+\sin{C}\le \frac{3\sqrt{3}}{2}$.

Proof: $\sin$ is a concave function from $0\le \theta \le \pi$. Therefore we may use Jensen's inequality: $\frac{\sin{A}+\sin{B}+\sin{C}}{3}\le \sin{\left(\frac{A+B+C}{3}\right)}=\frac{\sqrt{3}}{2}$

Alternatively, we may use a method that can be called "perturbation". If we let all the angles $A,B,C$ be equal, we prove that if we make one angle greater and the other one smaller, we will decrease the total value of the expression. To prove this, all we need to show is if $0<A,B<180$, then $\sin(A+B)+\sin(A-B)<2\sin A$. This inequality reduces to $2\sin A \cos B<2\sin A$, which is equivalent to $\cos B<1$. Since this is always true for $0<B<180$, this inequality is true. Therefore, the maximum value of this expression is when $A=B=C=60$, which gives us the value $\sin{A}+\sin{B}+\sin {C}=\frac{3\sqrt{3}}{2}$.

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