Question:     Show that the harmonic mean is less than or equal to the arithmetic mean, and also less than or equal to the geometric mean, with equality if and only if $a=b$ ; ie., show that $$\frac{a+b}{2} \geq \sqrt{ab} \geq \frac{2ab}{a+b}$$   My attempt,  I've shown the first equality which is $$\frac{a+b}{2}-\frac{2ab}{a+b}=(a-b)^2\geq0$$   My first question, should I write $\frac{(a-b)^2}{2(a+b)}\geq0$ or just leave as $(a-b)^2\geq0$?  For the second,   $$\sqrt{ab}-\frac{2ab}{a+b}=\frac{\sqrt{ab}(a+b)-2ab}{a+b}$$  $$=\frac{a^{\frac{3}{2}} \sqrt{b}+\sqrt{a} b^{\frac{3}{2}}-2ab}{a+b}$$  $$=\frac{\sqrt{ab}(a+b-2\sqrt{ab})}{a+b}$$  I stuck at here. Can anyone explain to me how to solve this question Thanks in advance.  **I found other similar questions being asked in this site. But my question includes harmonic  which is not a duplicate.
123097
$ m_{a} = \frac{1}{2} (a + b) $   $ m_{g} =  \sqrt{ab} $   $ \frac{1}{m_{h}} = \frac{1}{2}(\frac{1}{a} + \frac{1}{b})$   Attempted Solution:  I believe I have shown the first step, which is $ m_{a} \geq m_{g} $ but i'm stuck on showing $ m_{g} \geq m_{h} $  $ \frac{a + b}{2} &gt; \sqrt{ab} $  $ (\frac{a + b}{2})^2 &gt; (\sqrt{ab})^2 $  $\frac{a^2 + 2ab + b^2}{4} &gt; ab$  $a^2 + 2ab + b^2 &gt; 4ab $  $a^2 -2ab + b^2 &gt; 0 $  $(a - b)^2 &gt; 0 $   so if $a = b$ we have $(a -b)^2 \geq 0$  Therefore, $m_{a} \geq m_{g} $  Now, i'm not sure if that is correct or not, if not can you please guide me to the complete solution or if I am correct on this part, provide a solution for $m_{g} \geq m_{h} $
61943
Can anyone point me to any reference on the following inequality: Let $c&gt;0, a,b \in \Bbb{R}$ $$ \|a+b\|^2 \leq (1+c)\|a\|^2+(1+{\textstyle\frac{1}{c}})\|b\|^2 $$  I'm not even sure if "Bohr's Inequality" is this inequality's name, since I can't find it anywhere in the internet.  I'm having a bad day and I got stuck in this one, apparently. Any suggestions or hints on how to prove it will be appreciated.
102729
Prove: that $$(a^3+b^3)^2\leq (a^2+b^2)(a^4+b^4)$$ for all real numbers $a$ and $b$.   My attempt:  $(a^2+b^2)(a^4+b^4)=a^6+a^2 b^4+b^2 a^4+b^6\geq 4\cdot(a^6\cdot b^6\cdot a^2 b^4\cdot a^4 b^2)^{\frac{1}{4}}=4\cdot a^3\cdot b^3$, by AM-GM.  Now $(a^3+b^3)^2\geq 4a^3\cdot b^3$, by AM-GM. putting this above gives:$$(a^2+b^2)(a^4+b^4)\geq 4\cdot a^3\cdot b^3\leq (a^3+b^3)^2$$  But this don't gives good, please help, on what I did. How to improve solving Inequalities?
71521
If $a,b,c$ are real numbers such that $0\le a,b,c\le 1$, then $$\frac a{1+bc}+\frac b{1+ac}+\frac c{1+ab}\le 2.$$  I did the following: Take $f(x)=\frac {x^2}{x+p},\ (p=abc)$ which is convex for $0\le x\le 1$. Therefore $$\begin{align}\frac {f(a)+f(b)+f(c)}3\le f(\frac {a+b+c}3)\Rightarrow\frac {a^2}{a+abc}+\frac {b^2}{b+abc}+\frac {c^2}{c+abc}\le \frac {(a+b+c)^2}{(a+b+c)+3abc}.\end{align}$$ How do I prove that the RHS is less than or equal to 2?
25334
I have a math homework where it's being asked to prove that :  $$\forall a \geq 0,\sqrt{a}\leq\frac{1+a}{2}$$  However, I don't have any idea how I should start this one...  Any idea ?
88269
Show that $\frac c {1+c} \le \frac a {1+a} + \frac b {1+b}$ , for $c \le a+b$ and $a,b,c \ge 0$  So need to show $\frac c {1+c} \le \frac {a+b+2ab} {1+a+b+ab}$  We have $\frac c {1+c} \le \frac {a+b} {1+c}\le \frac {a+b+2ab} {1+c}$  So the whole thing reduces to showing $c \ge a+b+ab$ ?  Can anyone help me with this, I'm rubbish at inequalities...
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.
54416
Disclaimer: The statement may be false, but for now I'm operating under the assumption it's true and trying to prove it.   My workings:  I got a common denominator and expressed it as:  $$\frac{ab^2 + bc^2 + ca^2}{abc} \ge 3$$ $$ab^2 + bc^2 + ca^2 \ge 3abc$$  This is where I'm stuck. I feel as if I keep encountering this same result in various different problems and always am stumped on how to show it is true.   I tried to look at similar problems and it seems AM-GM inequality is the usual line of attack, so using that I reasoned, $$a^2 + b^2 + c^2 \ge 3abc$$, and now I need to establish $$ca^2 + ab^2 + bc^2 \ge a^2 + b^2 + c^2$$  However, this doesn't seem like it'd be true for general positive $a,b,c$.
77954
Let $a$ and $b$ be positive integers. Show that $\sqrt{2}$ always lies between $\dfrac{a}{b}$ and $\dfrac{a+2b}{a+b}$.  Please give the easy solution as possible.
84357
How to prove that $a^2+ab+b^2\ge 0$?  Obviously the squares are positive, but how can I be sure that $ab$ doesn't become too negative with a certain combination of $a$ and $b$?
42387
Kazakstan 2012 Suppose that $a, b \in\mathbb{R}$, and $a,b&gt;0$. If $\frac{1}{a}+\frac{1}{b}=2$ prove that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}.$ My idea : $a+b+\frac{1}{1+\sqrt{ab}}$ can be written as $ab(\frac{a+b}{ab}+\frac{1-\sqrt{ab}}{1-ab})=2ab+\frac{1-\sqrt{ab}}{ab-a^2b^2}$, but where  can I use the fact that $\frac{1}{a}+\frac{1}{b}=2?$
60534
If the $(\int_a ^b f(x))/(a-b)$ is the arithmetic average of all the values of $f(x)$ between $a$ and $b$, what is the expression representing the geometric average of all these values?
122098
Refer to the problem below (IMO 2009 Shortlist)     Let $a, b, c$ be positive real numbers such that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = a + b+ c$$   Prove that    $$\frac{1}{(2a+b+c)^2} + \frac{1}{(2a+b+c)^2} + \frac{1}{(2a+b+c)^2} \leqslant \frac{3}{16} $$    In the solution given by the  (Pg 16), it states that      Without loss of generality, we choose $$a +b+c = 1$$.    Thus, the problem becomes    $$ \frac{1}{(1 + a)^2} + \frac{1}{(1 + b)^2} +\frac{1}{(1 + b)^2} \leqslant \frac{3}{16}(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$$    Applying Jensen’s inequality to the function $$f(x) = \frac{x}{ (1+ x)^2}$$   ,which is concave for $0 ≤ x ≤ 2$   and increasing for $0 ≤ x ≤ 1$, we obtain   $$α \frac{a}{(1 + a)^2} + β \frac{b}{(1 + b)^2} + γ \frac{c}{(1 + c)^2} \leqslant (α + β + γ) \frac{A}{(1 + A)^2}$$   , where $A =\frac{αa + βb + γc}{α + β + γ}.$   Choosing $α = \frac{1}{a} , β =\frac{1}{b},$ and  $γ = \frac{1}{c}$   , we can apply the harmonic-arithmetic-mean inequality   $$A =\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} ≤ \frac{a + b + c }{3} = \frac{1}{3} &lt; 1$$   Finally we prove:   $$ \frac{1}{(1 + a)^2} + \frac{1}{(1 + b)^2} +\frac{1}{(1 + b)^2} \leqslant ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \frac{A}{(1 + A)^2} \leqslant ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c})\frac{A}{(1 + \frac{1}{3})^2} = \frac{3}{16}(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) $$   However,  (Pg 4) differs by having $a +b+c=3$ and setting $f(x) = \frac{1}{16x} - \frac{1}{(x+3)^2}$ and allowing Jensen to resolve the rest.   The questions are as follows;   How do you choose before hand which $a + b+ c =$ to choose?  Why particularly does the assumption differ between the 2 solutions? How does one know before hand the choosing of $α = \frac{1}{a} , β =\frac{1}{b},$ and  $γ = \frac{1}{c}$?  Where and how did $A =\frac{αa + βb + γc}{α + β + γ}$ come from?   Any help would be much appreciated.
49216
Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$,   Prove that $(a+b)(b+c)(c+a)\ge8$.   My attempt: By AM-GM inequality, we have  $$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$  and similarly  $$\frac{b+c}{2}\ge\sqrt {bc} \tag{2},$$ $$\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.$$  Multiplying $(1)$, $(2)$ and $(3)$ together we reach  $$(a+b)(b+c)(c+a) \ge8abc.$$  Now, I need to show that $abc = 1$.  Again by AM-GM inequality we have  $$\frac{a+b+c}{3}\ge\sqrt[3]{abc} \implies \frac{\frac{3}{abc}}{3}\ge\sqrt[3]{abc} \implies abc\le1$$    Now if we show somehow that $abc\ge1$, we are done and that's where I am stuck. Can someone please explain how to show that?  Other solutions to the above question are also welcomed.
44952
Let $a$, $b$, and $c$ be positive real numbers. Prove that  $$\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$$  Under what conditions does equality occur? That is, for what values of $a$, $b$, and $c$ are the two sides equal?
106250
Given Question is Prove that  $$\frac{b-a}{\sqrt{1-a^2}}&lt;\sin^{-1}b-\sin^{-1}a&lt;\frac{b-a}{\sqrt{1-b^2}}$$ when $0&lt;a&lt;b&lt;1$ using Mean Value Thoerem.  I considered $f(x)=\sin^{-1}x $ when $x\in(0,b)$ and used LMVT and established that $\frac{b}{\sqrt{a-c^2}}=\sin^{-1}b$ where $c \in (0,b)$.  But I don't know how to proceed after this. Please guide.
91462
Prove that the general arithmetic-geometric mean inequality \begin{equation*} (a_{1}a_{2}...a_{n})^\frac{1}{n}\leq\frac{a_{1}+a_{2}+...+a_{n}}{n} \end{equation*} holds for all $a_{i}$ positive real numbers.  I keep getting stuck half way. This is review material for me (which I feel like I should be getting easily, but that's not the case unfortunately).