If a belongs to(0,π/2) then √(x^2+x) + tan a/√(x^2+x) is greater than or equal to (x!=0,-1) is

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Aditya Gupta · Answer

let √(x^2+x) =y now simply apply the AM-GM inequality as follows: (y+tana/y)/2 >= root(tana) or (y+tana/y)>= 2 √tan a

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If a belongs to(0,π/2) then √(x^2+x) + tan a/√(x^2+x) is greater than or equal to (x!=0,-1) is

If a belongs to(0,π/2) then √(x^2+x) + tan a/√(x^2+x) is greater than or equal to (x!=0,-1) is

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If a belongs to(0,π/2) then √(x^2+x) + tan a/√(x^2+x) is greater than or equal to (x!=0,-1) is

If a belongs to(0,π/2) then √(x^2+x) + tan a/√(x^2+x) is greater than or equal to (x!=0,-1) is

1 Answers

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