How to prove this:     $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a}+ \frac{3\sqrt[3]{abc}}{a+b+c}\geq 4 $$   (Sorry I've asked this question on phone.)  I can understand the C-S step, but I've never learnt about the Schur inequality... I have googled it but I have no idea how to apply it here :/   I've done it by using harmonic mean to the Schur
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I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from  . The inequality which I tried to solve and I didn't manage to find a solution can be found in that pdf but I will write here to be more explicitly.  Exercise 1.3.4(a) Let $a,b,c$ be positive real numbers. Prove that $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4.$$ (b) For real numbers $a,b,c \gt0$ and $n \leq3$ prove that: $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$
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Suppose that $a$, $b$ and $c$ are positive real numbers and that $a \leq b + c$. By cross multiplying or otherwise, show that  $$\frac{a}{1 + a} \leq \frac{b}{1 + b} + \frac{c}{1 + c}$$  Anyone able to get this to work out? I cross multiplied them but am unable to manipulate it into an expression that verifies the statement.
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Prove that for real numbers $a_1 ,a_2 ,...,a_n &gt;0$ the following inequality holds $$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n }\leq 4\cdot \left(\frac{1}{a_1} +\frac{1}{a_2 } +...+\frac{1}{a_n} \right).$$
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Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) &gt; (a+b+c)(a^2+b^2+c^2)$.  I know the proof by subtracting LHS by RHS and then doing some arrangement.  But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) &gt;(a+b+c)(a^2+b^2+c^2)$, or an easy proof?
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If $a,b,c &gt; 0$ are such that $abc=1$, then   $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b+ c. $$   I would be pleased if you give me a hint. Thanks in advance.
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Let $a, b, c$ and $d$ be positive real numbers such that $a+b+c+d = 4$. Prove that $$\frac4{abcd} \geq \frac a b + \frac bc + \frac cd +\frac d a .$$   How can I approach this using only the ? Are there any other methods that do not involve the AM-GM inequality?
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I'd like to prove that $$\frac{(a-b)^2}{8a}\leqslant\frac{a+b}{2} - \sqrt{ab}\leqslant\frac{(a-b)^2}{8b}$$ when $ 0&lt;b\leqslant a\,$. I have been brushing up on inequalities and I cannot figure this one out. I have searched this site, and found this  , but haven't seen how to apply it. I have tried to find a factor that I can multiply the entire inequality with that will allow me to simplify the middle term. So far no luck there. I have also tried assuming that $b &gt; a$ and looked for a contradiction to either of the terms individually. Does anyone have any hints that can get me somewhere?
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I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives   $$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$   and on $abc$ which gives   $$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$   Since both inequalities have the same righthand side, I have tried to deduce something about the lefthand sides, but to no avail. Can somebody help me, please? I am sure it is something simple I have missed.
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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.
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Prove the inequality $$\frac{1}{\sqrt{1}+ \sqrt{3}}  +\frac{1}{\sqrt{5}+ \sqrt{7} }+......... +\frac{1}{\sqrt{9997}+\sqrt{9999}} &gt; 24$$   My work:   Rationalizing the denominator gives   $$\frac{\sqrt{3}-1}{2}  +\frac{\sqrt{7}-\sqrt{5}}{2}+......+\frac{\sqrt{9999}-\sqrt{9997}}{2} .$$  Now by taking two as common and separating the positive and negative terms gives  $$\frac{1}{2} [ \{\sqrt{3} +\sqrt{7}+\dots +\sqrt{9999}\} - \{1+\sqrt{5} +\dots+\sqrt{9997}\}].$$  Can we do like this please suggest. Thanks.
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Let $m_a, m_b, m_c - $ are median of triangle $ABC$.   Prove that:   $$ m_a + m_b + m_c \le 4R + r $$   I used   $$m_a^2=\frac{2b^2+2c^2-a^2}{4}$$ $$R=\frac{abc}{4R}$$ $$r=\frac{2S}{a+b+c}$$ and $R \ge 2r$
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For positive real numbers $a,b,c$ with $abc = 8$ prove that   $$ \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} + \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}} + \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}. $$   Can we prove this by Cauchy-Schwarz or Jensen's inequality? If not how?
18735
Prove that, $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it. Typically, I am actually looking for a little advanced and elegant solution.    EDIT: $a,b,c,d&gt;0$
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This is supposed to be an application of AM-GM inequality.     if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$   First of all,  $a^2+b^2+c^2\ge 3$  by a direct application of AM-GM.Also,we have  $a^2+b^2+c^2\ge ab+bc+ca$  Next,we consider the expression  $(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$  but that hardly helps.I know that  $3(a^2+b^2+c^2)\ge (a+b+c)^2$  From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,  $a^3+b^3+c^3\ge a^2b+b^2c+c^2a$  $(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$ A hint will be appreciated at this point.
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Prove that if a,b.c are distinct positive integers that $$a^4+b^4+c^4&gt;abc(a+b+c)$$ My attempt:  I used the inequality A.M>G.M to get two inequalities First inequality $$\frac{a^4+b^4+c^4}{3} &gt; \sqrt[3]{a^4b^4c^4}$$  or $$\frac{a^4+b^4+c^4}{3} &gt; abc \sqrt[3]{abc}$$ or new -- first inequality $$\frac{a^4+b^4+c^4}{3abc} &gt; \sqrt[3]{abc}$$  second inequality: $$\frac{a+b+c}{3} &gt; \sqrt[3]{abc}$$ I am seeing the numbers in the required equation variables here but am not able to manipulate these to get the inequality I want?? Please direct me on which step should I take after this??
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I want to prove the following inequality :      $$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab,  $$   $a,b,c,d \in \mathbb{R}_{+} .$    In my book, at the answers chapter the author uses AM $\geq$ GM, but I haven't any idea how I can use that.   Thanks :)
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How can one show that for triangles of sides $a,b,c$ that   $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} &lt; 2$$  My proof is long winded, which is why I am posting the problem here.   Step 1: let $a=x+y$, $b=y+z$, $c=x+z$, and let $x+y+z=1$ to get   $\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}&lt;2$  Step 2: consider the function $f(x)=\frac{1-x}{1+x}$, and note that it is convex on the interval (0,1), so the minimum of   $\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}$  is reached when the function takes the extreme points. i.e. $x=y=0, z=1$.    But going back to the fact that this is a triangle, we note that $x=y=0 \implies a=0$ which is not possible, so the inequality is strict.