2026/050) Show that if a,b,c are sides of a right angled triangle and c is the hypotenuse then $a^n+b^n \lt c^n$ for $c \gt 2$

We shall prove it by Principle of  Mathematical Induction

The base step is n = 3 and n = 4

Because  it is right angled triangle and c is the hypotenuse we have

$a^2 + b^2 = c^2\cdots(1)$

Further as  

$a < c$

Multiplying both sides by  a^2 we have

$a^3 < ca^2\cdots(2)$

Similarly 

$b^3 < cb^2\cdots(3)$

Adding (2) and (3) we get

$a^3 + b^3 < ca^2 + cb^2$

or 

$a^3+b^3 < c(a^2+b^2)$

or $a^3 + b^3 <c^3$ using (2)

Additionally

Squaring (1) we get 

$a^4 + b^4 + 2ab = c^4$

And hence dropping 2ab(which is positive) from the left 

$a^4 + b^4 < c^4$    

So we have proved the base step

For induction step we shall not move from n to n+1 but ftom n to n+ 2

We have Let t be true upto k = n

from k = n- 1  (not k) we shall prove for n+ 1

we have

$a^{k-1} < c^{k-1}$

Multiplying by a^2 on both sides we get

$a^{k +1} < c^{k-1}a^2\cdots(4)$

Similarly

 $b^{k +1} < c^{k-1}b^2\cdots(5)$

from (4) and (5) we get 

 $a^{k +1}  + b^{k+1} < c^{k-1}a^2 + c^{k-1}b^2$

Or  

 $a^{k +1}  + b^{k+1} < c^{k-1}(a^2 + b^2)$

or   $a^{k +1}  + b^{k+1} < c^{k-1}c^2)$

or  $a^{k +1}  + b^{k+1} < c^{k+1}$

So the induction step is complete

We have proved for 3 and 4 ad from n-1 to n+ 1 and hence for all above 2

As we have proved base step and induction step proof is complete

 

 

 

 

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