A=(1+x2m)(1+y2n)xmyn=(x−m+xm)(y−n+yn)1 2xm+y−m≥(xn+1,x−m)21⇒xn++x−n≥2 In the same way, yn+yn≥2 Then ,(xn+x−n)(y−n+yn)≥4 ⇒(x"1 +x−m)(y−n+yn)1≤41
Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression x^ m y^ n (1+x^2 ~m ) (1+y^2 n ) is :
JEE Main 2019 — Mathematics Calculus
Let x,y be positive real numbers and m,n positive integers. The maximum value of the expression (1+x2 m)(1+y2n)xmyn is :
Held on 11 Jan 2019 · Verified 6 Jul 2026.
Options
- A
1
- B
21
- C
41
- D
6mnm+n
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