How to prove that $\sqrt{2011} + \sqrt{2013} + \sqrt{2015} + \sqrt{2017} < 4\sqrt{2014}$ without using calculator?
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2 Answers
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The function $f(x)=\sqrt{x}$ is strictly concave for $x>0$ so by Jensen's inequality, we have $$ \frac{1}{4}f(2011)+\frac{1}{4}f(2013)+\frac{1}{4}f(2015)+\frac{1}{4}f(2017)<f\left(\frac{1}{4}(2011+2013+2015+2017)\right) $$ which is $\sqrt{2014}$.
answered Mar 30, 2015 at 20:03
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Hint:
$$\sqrt{k-r}+\sqrt{k+r}<2\sqrt{k},\ \ \ \forall k\ge 0, k\ge |r|, r\neq 0$$
Since:
$$\iff 2k+2\sqrt{k^2-r^2}<4k\iff 2\sqrt{k^2-r^2}<2k\iff k^2-r^2<k^2\iff -r^2<0$$
answered Mar 30, 2015 at 19:59