I would like to prove convexity of $\frac{1}{x}$. It can be proved by using second derivative but I want without using second derivative. Can someone help me?
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Have you tried to use Jensen's inequality? – Michael Galuza Jul 17 '15 at 07:43
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Find relationship between $\frac{1}{a}+\frac{1}{b}$ and $\frac{2}{a+b}$ – Brian Cheung Jul 17 '15 at 07:43
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Thank you would you please more explain? – Ali Jul 17 '15 at 07:46
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Use the definition of convexity. – Axel Jul 17 '15 at 08:02
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It is easy to see that $\frac{a+b}{ab}<\frac{2}{ab}$ by assuming $a<1,b<1$ then? – Ali Jul 17 '15 at 08:05
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Show that $\frac{1}{2} \left(\frac{1}{a} + \frac{1}{b}\right) \geq \frac{1}{\frac{a+b}{2}}$ for any $a>0, b>0$. – Axel Jul 17 '15 at 08:12
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We can use this criterion. Now pick $x>0$, $y>0$ and remark that $$ f \left( \frac{x+y}{2} \right) \leq \frac{f(x)+f(y)}{2} $$ is equivalent to $$ \frac{1}{2} \left( \frac{1}{x}+\frac{1}{y} \right) \geq \frac{1}{\frac{x+y}{2}}. $$ By some algebraic manipulation, this is in turn equivalent to $(x+y)^2 \geq 4xy$, or $x^2-2xy +y^2 \geq 0$. Of course this is always true, since $x^2-2xy+y^2= (x-y)^2$.
