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

Answer

Verified

358.5k+ views

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.

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

â€¢

Total views: 358.5k

â€¢

Views today: 6.58k