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Show that ${({x^m} + {y^m})^n} < {({x^n} + {y^n})^m}$, if $m > n$.

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Hint: Here we will verify the polynomial equation ${({x^m} + {y^m})^n} < {({x^n} + {y^n})^m}$ by substituting the values for m and n satisfying the given conditions.

Complete step-by-step answer:
We have to show ${({x^m} + {y^m})^n} < {({x^n} + {y^n})^m}$, if $m > n$
Let m=2 and n=1 (as 2>1)
Now take L.H.S
$ \Rightarrow {({x^m} + {y^m})^n} = {({x^2} + {y^2})^1} = {x^2} + {y^2}$
Now take R.H.S
$ \Rightarrow {({x^n} + {y^n})^m} = {({x^1} + {y^1})^2} = {x^2} + {y^2} + 2xy$
From this clearly we can say that
  {x^2} + {y^2} < {x^2} + {y^2} + 2xy \\
   \Rightarrow L.H.S < R.H.S \\
   \Rightarrow {({x^m} + {y^m})^n} < {({x^n} + {y^n})^m} \\
Hence proved.

Note: Whenever we face such a type of problem, always put the smallest integer value in place of m and n satisfying the given condition, then simplify and verify it.
Last updated date: 15th Sep 2023
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