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

Last updated date: 28th Mar 2023
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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.

We have to show ${({x^m} + {y^m})^n} < {({x^n} + {y^n})^m}$, if $m > n$
$\Rightarrow {({x^m} + {y^m})^n} = {({x^2} + {y^2})^1} = {x^2} + {y^2}$
$\Rightarrow {({x^n} + {y^n})^m} = {({x^1} + {y^1})^2} = {x^2} + {y^2} + 2xy$
${x^2} + {y^2} < {x^2} + {y^2} + 2xy \\ \Rightarrow L.H.S < R.H.S \\ \Rightarrow {({x^m} + {y^m})^n} < {({x^n} + {y^n})^m} \\$