alfa problems

$Let\ us\ use\ A.M-H.M:$ $\dfrac{\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+...\dfrac{1}{a_{2015}}}{2015}\geq\dfrac{2015}{a_1+a_2+a_3+....a_{2015}}$ $But,a_1+a_2+a_3+....a_{2015}=1$ $\Longrightarrow\ \dfrac{\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+...\dfrac{1}{a_{2015}}}{2015}\geq2015$ $\Longrightarrow\ \dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+...\dfrac{1}{a_{2015}}\geq2015^{2}$ $\Longrightarrow\ answer=\sqrt{2015^{2}}=2015.$ $A.M-H.M\ works\ this\ way:\dfrac{(a_1+a_2+...a_n)}{n}\geq\dfrac{n}{\dfrac{1}{a_1}+\dfrac{1}{a_2}+...\dfrac{1}{a_n}}$

we can used Jenson's inequality

Let $y\quad =\quad f\left( x \right) \quad =\quad \cfrac { 1 }{ x } \quad \quad \quad \quad \quad x\quad \in \quad (0,1)$ .

Pic

Consider 2015 point's on the curve So after joining them an polygon is formed whose Centroid is G and which lies inside the polygon , and a point P which lies on the curve with same x-coordinate as that of centroid So

${ y }_{ G }\quad \ge \quad { y }_{ p }\\ \\ \frac { \sum { f\left( { x }_{ i } \right) } }{ n } \quad \ge \quad f\left( \cfrac { \sum { { x }_{ i } } }{ n } \right) \\ \\ \sum { f\left( { x }_{ i } \right) } \quad \ge \quad nf\left( \cfrac { 1 }{ n } \right) \quad \quad \quad (put\quad n\quad =\quad 2015)\\ \\ \sum { f\left( { x }_{ i } \right) } \quad \ge \quad { 2015 }^{ 2 }\\ \\ \boxed { k\quad =\quad 2015 }$ .

I did this through a simple application of Titu's Lemma. The key is to treat all the numerators as if they were the square of 1, rather than just 1. Easiest solution, in my opinion.