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Consider function $$f(x)=\frac{x^{n_{1}}}{1-x}+\frac{(1-x)^{n_{2}}}{x},x\in(0,1)$$ where $n_{1}$ and $n_2$ are some fixed positive integers.

My question: Is $f(x)$ convex for any fixed $n_1$ and $n_2$?

The second derivation of function $f$ is very complex, so I wish there exists other method to verify convex property.

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Yeah, you can always use the definition ;-) – dtldarek Apr 22 '12 at 12:51
up vote 4 down vote accepted

A method to show f is convex is to show $f''(x)>0$. Do the two terms separately, reduce to determining the sign of a quadratic polynomial in the numerator.

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Thank you very much! I work out it follow your idea. – Wayson Kong Apr 22 '12 at 14:26

In mathematics, a real-valued function defined on an interval is called convex (or convex downward or concave upward) if the graph of the function lies below the line segment joining any two points of the graph. Equivalently, a function is convex if its epigraph (the set of points on or above the graph of the function) is a convex set. More generally, this definition of convex functions makes sense for functions defined on a convex subset of any vector space.according to wikipedia

A real valued function f : X → R defined on a convex set X in a vector space is called convex if, for any two points x1,x2 in X and any t belongs [0 1] we have $f(t*x1+(1-t)*x2)<=(t*f(x1)+(1-t)f(x2))$ now let's take $n1$ and $n2$ some fixed values,let say 5 and 10,and try it

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Thank you for your answer. – Wayson Kong Apr 22 '12 at 12:39
you are welcome, good lucks – dato datuashvili Apr 22 '12 at 12:46

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