# How to prove $f(x,y)=\sqrt {xy}$ is concave?

How can I prove (preferably elegantly) that $f(x,y)=\sqrt {xy}$ where $x≥0$ and $y≥0$ is concave in both $x$ and $y$?

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What's wrong with computing the Hessian? – Willie Wong Jul 2 '12 at 9:47
A professor of mine told me, some years ago, that most mathematicians usually give a wrong answer to the question "for what values of $p$ and $q$ is the map $(x,y) \mapsto x^p y^q$ convex?" In general, the only good answer is via the second derivative. – Siminore Jul 2 '12 at 9:55

Take $(x_0, y_0)$ and $(x_1,y_1)$ and then you need to show $f( t( x_0 , y_0) + (1-t) (x_1 , y_1) ) \geq t f(( x_0 , y_0)) +(1-t) f((x_1 , y_1) )$ which by squaring it becomes $$(t x_0 +(1-t)x_1)( t y_0 +(1-t)y_1) \geq ( t \sqrt{ x_0 y_0} + (1-t) \sqrt{x_1 y_1 })^2$$ which hold trivially by Cauchy-Schwarz.