Let $a, b>0$. Prove that:
$$\left ( 1+ \frac{a}{b} \right )^{2014}+ \left ( 1+ \frac{b}{a} \right )^{2014}\geq 2^{2015}$$
Solution
Using the AM-GM inequality we have successively:
$$\begin{aligned}
\left ( 1+ \frac{a}{b} \right )^{2014}+ \left ( 1+ \frac{b}{a} \right )^{2014} &\geq 2\sqrt{\left [ \left ( 1+ \frac{a}{b} \right )\left ( 1+ \frac{b}{a} \right ) \right ]^{2014}} \\
&= 2\sqrt{\left ( \frac{a}{b}+ \frac{b}{a}+ 2\right )^{2014}}\\
&= 2\sqrt{\left [ \left(\sqrt{\frac{a}{b}}+ \sqrt{\frac{b}{a}} \right)^2\right ]^{2014}}\\
&= 2\left ( \sqrt{\frac{a}{b}}+ \sqrt{\frac{b}{a}} \right )^{2014}\\
&\geq 2^{2015}
\end{aligned}$$
The exercise can also be found in mathematica.gr
$$\left ( 1+ \frac{a}{b} \right )^{2014}+ \left ( 1+ \frac{b}{a} \right )^{2014}\geq 2^{2015}$$
Solution
Using the AM-GM inequality we have successively:
$$\begin{aligned}
\left ( 1+ \frac{a}{b} \right )^{2014}+ \left ( 1+ \frac{b}{a} \right )^{2014} &\geq 2\sqrt{\left [ \left ( 1+ \frac{a}{b} \right )\left ( 1+ \frac{b}{a} \right ) \right ]^{2014}} \\
&= 2\sqrt{\left ( \frac{a}{b}+ \frac{b}{a}+ 2\right )^{2014}}\\
&= 2\sqrt{\left [ \left(\sqrt{\frac{a}{b}}+ \sqrt{\frac{b}{a}} \right)^2\right ]^{2014}}\\
&= 2\left ( \sqrt{\frac{a}{b}}+ \sqrt{\frac{b}{a}} \right )^{2014}\\
&\geq 2^{2015}
\end{aligned}$$
The exercise can also be found in mathematica.gr
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