If $a, b&gt;0$ and n is a positive integer, prove that: $(1+\frac{a}{b})^n+(1+\frac{b}{a})^n\ge 2^{n+1}$ I solved it in the following way: $(1+\frac{a}{b})^n+(1+\frac{b}{a})^n \ge (2\sqrt{\frac{a}{b}})^n+(2\sqrt{\frac{b}{a}})^n=2^n((\frac{a}{b})^n+(\frac{b}{a})^n)$ It is enough if $(\sqrt{\frac{a}{b}})^n+(\sqrt{\frac{b}{a}})^n\ge2$ If $n=1$: $\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\ge2$ $\frac{a}{b}+\frac{b}{a}+2\ge 4$ which holds true. If it holds true for $n=k$: If $b\ge a$: $(\sqrt{\frac{a}{b}})^k+(\sqrt{\frac{b}{a}})^k$ $(\sqrt{\frac{a}{b}})^{k+1}+\sqrt{\frac{a}{b}}*(\sqrt{\frac{b}{a}})^k\ge 2$ $(\sqrt{\frac{a}{b}})^{k+1}+(\sqrt{\frac{b}{a}})^{k-1}\ge 2$ And since $b\ge a$ We have that the original proposition holds true for k+1. Similarly, we prove that it is true if $a\ge b$. Hence because of induction, it holds true. Could you please show some other methods of solving this question easier?
58156
Let $a$ and $b$ be positive real numbers and let $n$ be a natural number prove that   $$\left({1+\frac ab}\right)^n+\left(1+\frac ba\right)^n\ge2^{n+1}.$$
61352
If $n$ be a positive integer greater than $1$, then prove that $$2^n&gt;1+n\sqrt{2^{n-1}}$$    I found this problem under $AM \ge GM$ chapter. Help me to solve this problem using  $AM \ge GM$. Thanks in advance.
1706
I am having trouble with the following proof:     For every positive integer $n$: $$1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$$   My work: I have tried to add $\frac{1}{(k+1)^2}$ to $2-\frac{1}{k}$ in the inductive step and reduce it down to $2-\frac{1}{k+1}$ but cannot do so. I am beginning to think that my entire approach is wrong.
122721
I need to show that for $x&gt;0$, $1+\frac{x}{2} \ge \sqrt{1+x} \ge 1+\frac{x}{2} - \frac{x^2}{8} $.  I used the geometric/arithmetic mean inequality to show that $1+\frac{x}{2} \ge \sqrt{1+x}$ is indeed true. My issue lies in the second part of this.  In trying to show that $\sqrt{1+x} \ge 1+\frac{x}{2} - \frac{x^2}{8} $, I determined that this is a partial taylor series expansion and I attempted to play around with both inequalities to show that this is true but to no avail.  I would appreciate any help in solving this. Ideally I would like to do this analytically but any solutions using more complicated theorems are welcome.
40943
If $a,b,c$ are positive real numbers, not all equal, prove that  $$\dfrac{bc}{b+c}+\dfrac{ca}{c+a}+\dfrac{ab}{a+b}&lt;\dfrac{a+b+c}{2}$$  Progress: $\frac{bc}{b+c}-\frac{b+c}{2}=$
114993
I was trying to prove this inequality using induction, but couldn't do.  Question: Suppose $a$ and $b$ are real numbers with $0 &lt; b &lt; a$. Prove that if $n$ is a positive integer, then:  $$a^n-b^n \leq na^{n-1}(a-b)$$
5998
Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} &gt; \frac{13}{24}$ for $n&gt;1,n\in\Bbb N$  To solve it I used induction but it is leading me nowhere my attempt was as follows:  Lets assume the inequality is true for $n = k$ then we need to prove that it is true for $k+1$  so we need to prove $\frac1{k+2} + \frac1{k+3}+\cdots+\frac1{2(k+1)} &gt; 13/24$  I don't know where to go from here please help.
23022
If $a$ and $b$ are positive real numbers, then $a + b \geq 2 \sqrt{ab}$.   I know how to do the direct proof, but in this case, I want to try proving it by contradiction. I have tried manipulating the inequality $a + b &lt; 2 \sqrt{ab}$ after making the assumption that $a,b &gt;0$ to get a contadiction $a,b \leq 0$.  $\begin{align} a + b &amp;&lt; 2 \sqrt{ab} \\a^2-2ab+b^2 &amp;&lt; 0 \\(a-b)^2 &amp;&lt; 0\end{align}$  How do I show that $a,b \leq 0$?
67806
$$\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n} &gt; \ln 2$$  $n$ is positive integer.  Thank you !
3730
Let $a$, $b$ and $c$ be positive real numbers.   $(\mathrm{i})$ Prove that $4(a^3 + b^3) \ge (a + b)^3$.   $(\mathrm{ii})$Prove that $9(a^3 + b^3 + c^3) \ge (a + b + c)^3.$   For the first one I tried expanding to get $a^3 + b^3 \ge a^2b+ab^2$ but I'm not sure how to prove it.
99210
Prove that for all  $n \in \mathbb{N}$: $$2^{n-1}(a^n + b^n) \ge (a + b)^n$$ I used induction (for $k = 1,2,...n-1$), and ended up with the following equation:  $$(a + b)(a + b)^{n-1} \le (a + b)(a^{n-1} + b^{n-1})2^{n-2}$$  Now, I have a feeling I'm on the right track, but I'm not totally sure. Could someone give me a pointer one where I should go from here? Any help would be greatly appreciated. Thanks!
41505
Let $a$, $b$ and $c$ be positive numbers. Prove that:   $$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$   This inequality is strengthening of the following Vasile Cirtoaje's one, which he  created in 2005.     Let $a$, $b$ and $c$ be positive numbers. Prove that:   $$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$$   My proof of this inequality you can see here:   But this way does not help for the starting inequality.  A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is  $0.0000158...$.  I tried also to use Cauchy-Schwarz, but without success.  Also, I think the BW (see here  I tryed!) does not help.
48670
$a,b,c$ are positive real numbers such that, $a+b+c\ge abc$. Prove that $a^2+b^2+c^2\ge \sqrt{3}abc$    My work: I tried using Cauchy-Schwarz inequality to find that, $(a^2+b^2+c^2)(1^2+1^2+1^2)\ge (a+b+c)^2$ $(a^2+b^2+c^2)\ge \dfrac13(a+b+c)^2$ $\sqrt{(a^2+b^2+c^2)}\ge \dfrac{1}{\sqrt{3}}(a+b+c)\ge \dfrac{1}{\sqrt{3}}abc$ which is not what I need and neither I can use it to prove the required inequality. Please help.
95471
I just thought of this proof but I can't seem to get it to work.  Let $a,b,c&gt;0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$  Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and similarly for $b,c$. Hence it suffices to prove that $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}$$. From $a+b+c=1$ and $a,b,c&gt;0$ we have $0&lt;a,b,c&lt;1$, so we have $$\frac{a}{1-a}=a+a^2+a^3+...$$ and similarly for $b,c$, so it suffices to prove that $$\sum_{cyc} a+\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{3}{2}$$, or equivalently (by $a+b+c=1$) $$\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{1}{2}$$, where $\sum_{cyc} a=a+b+c$ similarly for $\sum_{cyc}a^n=a^n+b^n+c^n$. Here I get stuck. For example, doing the stuff below yields a weak inequality, because of too many applications of the $a^2+b^2+c^2\ge (a+b+c)^2/3$ inequality.  "stuff below": Now, from $0&lt;a&lt;1$ we have $a^3&gt;a^4$, $a^5&gt;a^8$, $a^7&gt;a^8$, $a^9&gt;a^{16}$, and so on, so it suffices to prove that $$\sum_{cyc} a^2+2\sum_{cyc} a^4+4\sum_{cyc} a^8+8\sum_{cyc} a^{16}+...\ge \frac{1}{2}$$, or, multiplying by 2, $$2\sum_{cyc} a^2+4\sum_{cyc} a^4+8\sum_{cyc} a^8+...\ge 1$$, which by a simple inequality (i.e. recursively using $a^{2^n}+b^{2^n}+c^{2^n}\ge \frac{(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2}{3}$) is equivalent to $$2(1/3)+4(1/3)^3+8(1/3)^7+...+2^n(1/3)^{2^n-1}+...\ge 1$$. But then *%281%2F3%29%5E%282%5Ei-1%29 and so we're screwed.
97993
I was asked to prove an inequality: For any $n$ positive numbers $\{a_i\}$ with $a_{1}a_{2}\cdots a_{n} = 1$ and $m \geq n-1$ be a non-negative integer, $a_1^m + a_2^m + \cdots + a_n^m \geq \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}$  After several attempts, I think I had to use induction on both $m$ and $n$. In particular, I fix an $n$ and claim that the inequality holds for every $m$, and use induction on $n$.  The case where $n = 2$ could be done by rewriting the inequality as $a_1^2 + a_2^2 \geq a_1 + a_2$ and assume  that $a_1 \leq 1 \leq a_2$.   In the inductive step, given $a_1^k + a_2^k \geq a_1 + a_2$ , by writing $a_1^{k+1} + a_2^{k+1} = (a_1^{k} + a_2^{k})(a_1 + a_2) - a_1^{k}a_2^{k} (a_1 + a_2)$, the result follows.  However, this method seems to fail for $n&gt;2$, as the right-hand side of the inequality becomes more complicated. Should I try not to prove it by induction?
19168
Let $a$ be a nonzero number and $m$ and $n$ be integers. Prove the following equality: $a^{m+n}=a^{m}a^{n}$  I'm not really sure what direction to go in. I'm not sure if I need to show for $n$ positive and negative separately or is there an easier way. Can you use induction on integers?  My attempt: 1) Base case $m=0$. Prove that $a^{m+n}=a^m a^n$. Is true.  2) Assume the result holds for $m$.  So I want to prove it holds for $m+1$.  So I know that $a^{m+n}=a^ma^n$. So does this imply that: $a^{m+n+1}= a^{m+1+n}= a^{m+1}a^n$? Am I going in the right direction? I'm not sure what to do next...
67186
I have to do proof of $(a+b)^n$ and $(a+b)^{n+1}$ with mathematical induction.  I finished the first one $(a+b)^n = a^n + na^{n-1}b+\dots+b^n$.  I however have trouble with the second one, I don't know what to start exactly. Any help/hints?