Question

In a right angled triangle ABC, right angled at B prove that \[{\left( {AB} \right)^3} + {\left( {BC} \right)^3} < {\left( {AC} \right)^3}\] ?

Answer 1

Hint: In the above question, we are given a right angled triangle ABC having a right angle at the vertex B. We have to prove that \[{\left( {AB} \right)^3} + {\left( {BC} \right)^3} < {\left( {AC} \right)^3}\] . In order to approach the solution, we can use the well-known theorem of a right triangle known as the Pythagoras theorem, which states that the square of the longest side, i.e. the hypotenuse is equal to the sum of the squares of the other two sides. Mathematically that is written as \[{\left( {AB} \right)^2} + {\left( {BC} \right)^2} = {\left( {AC} \right)^2}\] .

Complete step by step answer:
Given that, a right angled triangle \[\vartriangle ABC\] which is right angled at \[B\] , i.e. \[\angle B = 90^\circ \]
We have to prove that \[{\left( {AB} \right)^3} + {\left( {BC} \right)^3} < {\left( {AC} \right)^3}\] .
Since, the hypotenuse is the longest side of a right angled triangle, and here \[AC\] is the hypotenuse.
Hence,
\[ \Rightarrow AC > AB\]
And
\[ \Rightarrow AC > BC\]
Therefore,
\[ \Rightarrow 0 < \dfrac{{AB}}{{AC}} < 1\]
and
\[ \Rightarrow 0 < \dfrac{{BC}}{{AC}} < 1\]
Since they are less than one, so their cubes will be less than their squares.
Therefore,
\[ \Rightarrow {\left( {\dfrac{{AB}}{{AC}}} \right)^3} < {\left( {\dfrac{{AB}}{{AC}}} \right)^2}\] ...(1)
And
\[ \Rightarrow {\left( {\dfrac{{BC}}{{AC}}} \right)^3} < {\left( {\dfrac{{BC}}{{AC}}} \right)^2}\] ...(2)
Adding (1) and (2) , we can write
\[ \Rightarrow {\left( {\dfrac{{AB}}{{AC}}} \right)^3} + {\left( {\dfrac{{BC}}{{AC}}} \right)^3} < {\left( {\dfrac{{AB}}{{AC}}} \right)^2} + {\left( {\dfrac{{BC}}{{AC}}} \right)^2}\]
That gives,
\[ \Rightarrow \left( {\dfrac{{A{B^3}}}{{A{C^3}}}} \right) + \left( {\dfrac{{B{C^3}}}{{A{C^3}}}} \right) < \left( {\dfrac{{A{B^2}}}{{A{C^2}}}} \right) + \left( {\dfrac{{B{C^2}}}{{A{C^2}}}} \right)\]
Adding the numerators of the RHS, we get
\[ \Rightarrow \left( {\dfrac{{A{B^3}}}{{A{C^3}}}} \right) + \left( {\dfrac{{B{C^3}}}{{A{C^3}}}} \right) < \left( {{{\dfrac{{A{B^2} + BC}}{{A{C^2}}}}^2}} \right)\]
Using the Pythagoras theorem and putting \[{\left( {AB} \right)^2} + {\left( {BC} \right)^2} = {\left( {AC} \right)^2}\] in above equation we can write,
\[ \Rightarrow \left( {\dfrac{{A{B^3}}}{{A{C^3}}}} \right) + \left( {\dfrac{{B{C^3}}}{{A{C^3}}}} \right) < \left( {{{\dfrac{{AC}}{{A{C^2}}}}^2}} \right)\]
That gives,
\[ \Rightarrow \left( {\dfrac{{A{B^3}}}{{A{C^3}}}} \right) + \left( {\dfrac{{B{C^3}}}{{A{C^3}}}} \right) < 1\]
Now adding the numerators of LHS, we get
\[ \Rightarrow \left( {\dfrac{{A{B^3} + B{C^3}}}{{A{C^3}}}} \right) < 1\]
Therefore,
\[ \Rightarrow A{B^3} + B{C^3} < A{C^3}\]
Hence, proved.
Therefore, we have proved \[A{B^3} + B{C^3} < A{C^3}\].

Note:
In any triangle, say \[\vartriangle ABC\] , where A, B and C are the three vertices of the triangle and AB, BC and AC are the three sides of the triangles respectively, then the sum of any two sides of the triangle is always greater than the third side of that triangle. Also, the sum of all three angles of any triangle \[\vartriangle ABC\] is equal to a straight angle, i.e. \[180^\circ \] .