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### algebrainc proof for $\alpha\beta \leq \frac{\alpha^p}{p} + \frac{\beta^q}{q}$ for conjugate exponents p and q [duplicate]

I found a very informal proof (geometric one, for by taking $\alpha$ as $\beta$ in the x-axis and y-axis and showing rectangle area is smaller than area under the plots). Is there a proper algebraic ...
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### How to show that $uv \le \frac {u^p}{p}+ \frac {v^q}{q}$ where $\frac 1p + \frac 1q = 1$ [duplicate]

Let $u, v, p, q > 0$, and let $\dfrac 1p + \dfrac 1q = 1$. Then $$uv \le \dfrac {u^p}{p}+ \dfrac {v^q}{q}$$ Show that equality occurs iff $u^p = v^q$. This is problem $10$ from Ch. 6 in ...
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### Given $y_n=(1+\frac{1}{n})^{n+1}$ show that $\lbrace y_n \rbrace$ is a decreasing sequence

Given $$y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1.$$ Show that $\lbrace y_n \rbrace$ is a decreasing sequence. Anyone can help ? I consider the ...
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### Binomial theorem proof for rational index without calculus

I have tried to find a proof of the binomial theorem for any power, but I am finding it difficult. One can obviously prove the integer index case using induction, but all of the approaches for ANY ...
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### Prove that $xy \leq\frac{x^p}{p} + \frac{y^q}{q}$

OK guys I have this problem: For $x,y,p,q>0$ and $\frac {1} {p} + \frac {1}{q}=1$ prove that $xy \leq\frac{x^p}{p} + \frac{y^q}{q}$ It says I should use Jensen's inequality, but I can't figure ...
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### Proving that $\left(1 + \frac{x}{n}\right)^n, n \in \mathbb N$ is bounded

Please can you help me with the following question $$E(x) = \left\{ \left(1 + \frac{x}{n}\right)^n : n \in \mathbb N \right\}$$ Let a(x) = sup E(x) (least upper bound) without finding the sup of E(...
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### Proof that $\lim_{x\to \infty} (1+(x/n))^n = e^x$ Using Monotone Convergence Theorem

Use monotone convergence theorem to prove that $e(x):=\lim_{n\to\infty}(1+x/n)^n$ exists for all real $x$. Then show that $e(-x):=\lim_{n\to\infty}(1-x/n)^n=1/e(x)$. I'm struggling to use monotone ...
### Binomial theorem and the inequality $(1-e^{-x})^{\alpha}\leq (1-\alpha e^{-x})$ for $0<\alpha\le 1$
Assume $0<\alpha\leq 1$ and $x>0$. Does the following inequality hold? $$(1-e^{-x})^{\alpha}\leq (1-\alpha e^{-x})$$ I know that the reverse inequality holds if $\alpha\ge 1$.