Can somebody explain me how to use Holder's inequality and Minkowski's inequality. Can these inequalities be used to solve the problem that if $x, y, k$ are non negative reals then prove that $$\left(1+\frac{x}{y}\right)^k +\left(1+ \frac{y}{x}\right)^k \ge 2^{k+1}$$ Edit 1- My main question is about what Holder's inequality and minkowski's inequality are and how do we use them
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This inequality is solved here by AM-GM: Prove the inequality $({1+\frac{a}b})^n$ + $(1+\frac{b}a)^n$ $\geq$ $2^{n+1}$ – Sil Nov 16 '17 at 17:48
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My main question is about what Holder's inequality and minkowski's inequality are and how do we use them – Rohan Shinde Nov 17 '17 at 03:07
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If that is the case, why would you accept rather quickly answer that does not address the Holder/Minkowski either? Anyway it is better to ask just one specific question, otherwise it confuse people (I haven't marked it as a duplicate though). Also questions of type "How to use Holder inequality" are quite generic, this site contains many Holder/Minkowski inequalities examples, you can just look at them. – Sil Nov 17 '17 at 04:42
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I asked the question so as it could be solved by Holder's inequality but the easy application of AM GM inequality itself provided the answer. So I felt there was no need to use Holder's inequality when the question can be solved so easily and accepted the answer. And sorry for any inconvenience hence caused.And thanks for help – Rohan Shinde Nov 17 '17 at 05:17
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Your inequality is not true if you allow $x, y \in \mathbb{R}$. Take $x = 1, y = -1$:
$$\left(1 + \frac{x}y\right)^k + \left(1 + \frac{y}x\right)^k = 0^k + 0^k = 0 < 2^{k+1}$$
However, it can be proved with AM-GM inequality if you assume $x,y > 0$:
\begin{align}\left(1 + \frac{x}y\right)^k + \left(1 + \frac{y}x\right)^k &= \sum_{i=0}^k {k \choose i}\frac{x^k}{y^k} + \sum_{i=0}^k {k \choose i}\frac{y^k}{x^k} \\ &= \sum_{i=0}^k {k \choose i}\left(\frac{x^k}{y^k} + \frac{y^k}{x^k}\right) \\ &\ge \sum_{i=0}^k {k \choose i}\cdot 2\sqrt{\frac{x^k}{y^k} \cdot \frac{y^k}{x^k}}\\ &= 2 \sum_{i=0}^k {k \choose i}\\ &= 2\cdot 2^k\\ &= 2^{k+1} \end{align}
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