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

Show that (xm+ym)n<(xn+yn)m{({x^m} + {y^m})^n} < {({x^n} + {y^n})^m}, if m>nm > n.

Explanation

Solution

Hint: Here we will verify the polynomial equation (xm+ym)n<(xn+yn)m{({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 (xm+ym)n<(xn+yn)m{({x^m} + {y^m})^n} < {({x^n} + {y^n})^m}, if m>nm > n
Let m=2 and n=1 (as 2>1)
Now take L.H.S
⇒(xm+ym)n=(x2+y2)1=x2+y2\Rightarrow {({x^m} + {y^m})^n} = {({x^2} + {y^2})^1} = {x^2} + {y^2}
Now take R.H.S
⇒(xn+yn)m=(x1+y1)2=x2+y2+2xy\Rightarrow {({x^n} + {y^n})^m} = {({x^1} + {y^1})^2} = {x^2} + {y^2} + 2xy
From this clearly we can say that
x2+y2<x2+y2+2xy ⇒L.H.S<R.H.S ⇒(xm+ym)n<(xn+yn)m  {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.