Given $a$ , $b$ , $c \ge 0$ show that   $$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} \ge \frac{3}{4}.$$   I tried using Titu's lemma on it, resulting in   $$\frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2 + 3(ab + bc + ca)} $$  And I am stuck here.
120974
Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$ for all notnegative value of $x,y,z$.  I think that minimum value is $\frac{3}{4}$ when $x=y=z$ but I have no prove.
50944
Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$  I tried it to do using $AM \ge GM$ but don't know how to proceed.  Please help.
113438
Let $a,b,c,d$ be positive real numbers. Prove that $$\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c} \geq \dfrac{2}{3}.$$   I was thinking of trying $a \geq b \geq c$ since the inequality is cyclic. I am not sure how to use this, though, or if this would help simplify the inequality. It also doesn't look like I can really use AM-GM or Cauchy-Schwarz.
40499
If $a, b, c$ belong to positive real numbers then prove that  $$\frac{2}{b+c} +\frac{2}{c+a}+\frac{2}{a+b} \le \frac 1a +\frac 1b +\frac 1c.$$  I applied AM-GM inequalities on both sides but I cant even figure out what to do on left side.
58414
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 :)
99875
Let $a,b,c,d\in \mathbb{R}$ and $a+b+c+d=2$. Prove that $$\frac{a}{a^2-a+1}+\frac{b}{b^2-b+1}+\frac{c}{c^2-c+1}+\frac{d}{d^2-d+1}\le \frac{8}{3}.$$   We have $$\frac{a}{a^2-a+1}\le \frac{4}{3}a\Longleftrightarrow \frac{-a\left(2a-1\right)^2}{3\left(a^2-a+1\right)}\le 0,$$ (Right)  So $$LHS\le \frac{4}{3}\left(a+b+c+d\right)=\frac{8}{3}=RHS$$  When $a=b=c=d=\frac{1}{2}$  But also $a=b=c=1;d=-1$ i don't know how to solve it.
95480
Let $a,b,c,d&gt;0$, show that  $$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}$$  I know this is interesting inequality,and Mathlinks can't solution,Thank you for you help  also can see this:  today,when I see this secrets in inequalitys Volume 1,(Pham kim Hung) page 205  let $a,b,c,d&gt;0$ prove that $$\sum_{cyc}\dfrac{a^2}{b}\ge 2\sqrt{2}\sqrt[4]{a^4+b^4+c^4+d^4}$$  I can't prove it,can you help me? Thank you
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.
114960
let $a,b,c\in \mathbb{R}$, if such  $$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$ show that $$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$  Does this problem has nice methods? My idea:let $$(ca-b^2)(ab-c^2)+(bc-a^2)(ab-c^2)+(bc-a^2)(ca-b^2)=0$$ then I can't. Thanks
60943
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.
10195
If $a, b, c \in \mathbb{N}$, and $abc = 1$, prove that: $$S = \frac{2}{(a+1)^2+b^2+1} + \frac{2}{(b+1)^2+c^2+1} + \frac{2}{(c+1)^2+a^2+1} \le 1$$  Here is my try: $$\begin{align}\frac{2}{(a+1)^2+(b^2+1)} &amp; \le \frac{1}{\sqrt{(a+1)^2 \cdot (b^2+1)}}\\ &amp; = \frac{1}{(a+1) \cdot \sqrt{b^2+1}}\\ &amp; \le \frac{1}{(a+1) \cdot b}\\ &amp; \le \frac{1}{ab} \end{align}$$  So, $$S \le \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{a+b+c}{1}$$  Can you help me to continue it, please? Thanks!
90070
Let $x,y,z&gt;0$. Show that $$\sum_{cyc}(1-x)^2\ge \sum_{cyc}\dfrac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$$ Here's what I have done. The expression $$\sum_{cyc}(1-x)\left((1-x)-\dfrac{z^2(1+x)(1-y^2)}{(xy+z)^2}\right)\ge 0$$ or $$\sum_{cyc}(1-x)\cdot\dfrac{x^2y^2+y^2z^2+xy^2z^2+2xyz-2xz^2-2x^2yz-x^3y^2}{(xy+z)^2}\ge 0.$$ But this way does not help for the starting inequality.
40775
a,b,c,d $\in \mathbb{R}$  $a,b,c,d \gt 0$ and  $ c^2 +d^2=(a^2 +b^2)^3$ prove that  $$ \frac{a^3}{c} + \frac{b^3}{d} \ge 1$$  If I rewrite the inequation like $ \frac{a^3}{c} + \frac{b^3}{d} \ge \frac{c^2 +d^2}{(a^2 +b^2)^3}$ and manage to simplfy  it brings me nowhere. I try with  Cauchy-Schwarz Inequality but still can not solve it. I would very much appreciate it if anyone could help me
104983
$a _{1} = \frac{1}{2}, a _{2} = 1, a _{n}= \frac{1}{2} a_{n-1} + \sqrt{a _{n-2} }$  Show by the induction that it's $\le 4$ and also by the induction that $a_{n+1}-a_n \ge 0$  If it comes to first I don't know what to do and if it comes to second I tried many different ways but I couldn't prove it.
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: $a,b,c,d&gt;0$
93377
Please help me for prove this inequality: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) &gt; 0$$
37835
Question:  let $a,b,c,d\ge 0$,such $$a^2+b^2+c^2+d^2=4$$ show that $$\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{d+3}+\dfrac{d}{a+3}\le 1$$  My try: By Cauchy-Schwarz inequality,we have $$\sum_{cyc}\dfrac{a}{b+3}\le\sqrt{(\sum_{cyc}a^2)\left(\sum_{cyc}\dfrac{1}{(b+3)^2}\right)}$$ then we have only prove this $$\sum_{cyc}\dfrac{1}{(a+3)^2}\le \dfrac{1}{4}?$$ This is  not true,in fact, we have $$\sum_{cyc}\dfrac{1}{(a+3)^2}\ge \dfrac{1}{4}?$$ because we have $$\dfrac{1}{(a+3)^2}\ge\dfrac{5-a^2}{64}$$ this is true because $$\Longleftrightarrow \dfrac{(a-1)^2(a^2+8a+19)}{64(a+3)^2}\ge 0$$ so $$\sum_{cyc}\dfrac{1}{(a+3)^2}\ge\sum_{cyc}\dfrac{5-a^2}{64}=\dfrac{1}{4}$$ can see:  This methods is from: can  see:  By the way I have see this three variable inequality  let $a,b,c$ be non-negative numbers such that $$a^2+b^2+c^2=3$$  show that $$\dfrac{a}{b+2}+\dfrac{b}{c+2}+\dfrac{c}{a+2}\le 1$$  proof: By expanding,the inequality becomes $$ab^2+bc^2+ca^2\le abc+2$$ without loss of generality,assume that  $$\min(a,b,c)\le b\max(a,b,c)$$ then \begin{align*} 2-ab^2-bc^2-ca^2+abc&amp;=2-ab^2-b(3-a^2-b^2)-ca^2+abc\\ &amp;=(b^3-3b+2)-a(b^2-ab+ca-bc)\\ &amp;=(b-1)^2(b+2)-a(b-a)(b-c)\ge 0 \end{align*} Equality occurs for $(a,b,c)=(1,1,1)$ and also for $(a,b,c)=(0,1,\sqrt{2})$ or any cyclic permutation.  so my Four-inequality  variable inequality,can use this methods $$\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{d+3}+\dfrac{d}{a+3}\le 1$$ $$\Longleftrightarrow a^2cd+3a^2c+3a^2d+9a^2+ab^2d+3ab^2+abc^2-abcd+3ac^2+9ac+3b^2d+9b^2+3bc^2+bcd^2+3bd^2+9bd+9c^2+3cd^2+9d^2-81\le 0$$ $$\Longleftrightarrow a^2cd+3a^2c+3a^2d+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45$$ $$\Longleftrightarrow a^2(cd+3c+3d)+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45$$ $$\Longleftrightarrow (4-b^2-c^2-d^2)(cd+3c+3d)+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45$$  Then  I can't  Thank you  very  much