Let $a,b,c $ be positive reals such that $abc=1$. Find the minimum value of $$Q = \frac{1}{a^3(b+c)}  + \frac{1}{b^3(a+c)} + \frac{1}{c^3 (a+b)} $$  My attempt:   Let $P = 2(ab + ba + bc)$.  Then $PQ &gt; \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} \geq  3 $, by AM-GM inequality.   Also, $P = 2(ab + ba + bc) = 2(\dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a}) \geq 6 $, by AM-GM inequality.  $P(1,1,1) = 6 $ and $Q(1,1,1) = \frac{3}{2}.$ Could anyone advise me how to show absolute minimum of $Q$ is $\frac{3}{2}$ ?  Hints will suffice, thank you.
25531
$a,b,c$ are positive reals with $abc = 1$. Prove that    $$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$$   I try to use AM $\ge$ HM.  $$\frac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)}}3\ge \frac{3}{a^3(b+c)+b^3(a+c)+c^3(a+b)}$$ Then how I proceed.
68034
If $a$,$b$ and $c$ positive real numbers such that $a+b+c=1$, prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$. I have tried several methods to solve this,but can't get any result. Any idea?
20703
let $a,b,c$ be postive real numbers ,and such $$ab+bc+ac=3$$   show that   $$\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge 1$$   This problem is from Secrets In Inequalities volume 1  page 30,example 1.24. the comment.the  author this case is a bit more diffcult,But this author can't post solution.Thank you
54078
Let $x,y,z&gt;0$ and $x+y+z=1$, then find the least value of  $${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$$  I tried various ways of rearranging and using AM > GM inequality. But I couldn't get it. I am not good at inequalities. Please help me.  I wrote $x$ as $1-(y+z)$ and I took $x+y$ as $a$ and the others as $b$ and $c$. And I am trying.
81550
Prove that the following inequality holds   $$a^a b^b c^c\ge (abc)^{\frac{a+b+c}{3}}$$   if $a,b,c$ are positive.   I'm not sure how to handle these kinds of powers. Are there any "famous" but not so advanced inequalities that involve this kinds of expressions?  I was trying to do something like ordering them without lost of generality because it's symmetric, but I get nowhere and I just make the thing even more complicated.  I'm asking for solution rather than hints since this is my first time encountering these kinds of problems.
58658
if $a,b,c$ are non zero positive reals prove $$\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$$ I have used A.M  G.M  inequality as follows:  $$a^3+b^3+c^3 \ge 3abc$$ adding $abc$ both sides we get  $$a^3+b^3+abc \ge 4abc-c^3=c(4ab-c^2)$$ so  $$\frac{1}{a^3+b^3+abc} \le \frac{1}{c(4ab-c^2)}$$ But since $a,b,c$ are positive reals  $$4abc-c^3 &lt; 4abc$$ so  $$\frac{1}{4abc-c^3} \gt \frac{1}{4abc}$$  but i am unable to proceed here
76815
Let $a,b,c$ be positive real numbers satisfying $$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=16,$$ then $a+b+c=3$.  I thought about the application of the AM-GM-HM inequality, but in vain. I also thought about splitting 16 into factors and comparing but went nowhere. Any ideas. Thanks beforehand.
113922
I can't see any ways I could use the fact that $a+b+c=1$. I've tried solving the inequality using various AM-GM inequalities, but I just can't make it. I need some help. Thanks.
23115
$(i)$ $a&gt;0,b&gt;0,p&gt;0,q&gt;0$   $(ii)$ $\frac{1}{p}+\frac{1}{q}=1$     How can we prove it?   ${a}{b}≤\frac{a^p}{p}+\frac{b^q}{q}$
35089
Is $a,b,c$ are three different positive integers such that :  $$ ab+bc+ca\geq 107 $$  Then what is the minimum value of $a^3+b^3+c^3-3abc$.  I expanded this expression to $(a+b+c)((a+b+c)^2-3(ab+bc+ca))$ and tried to find the minimum value of $(a+b+c)$ by AM-GM inequalities but the problem is that the minimum value occurs at scenario where $a=b=c$, which is ommitted by assumptions of this question. I'd appreciate some help.
60947
Let a,b,c be positive real number, such that  $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that :  $\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2} \leq \frac{3}{16}$  Can anyone help me how to deal with it?
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.
913
Suppose $a_1, a_2,\dots , a_n$ are $n$ positive real numbers with $a_1a_2 \dots a_n = 1$.   Then what is the minimum value of $(1 + a_1)(1 + a_2). . .(1 + a_n)$ ?    I think $(1 + a_1)(1 + a_2). . .(1 + a_n)$ takes its minimum value when $a_1=a_2=\dots=a_n=1$ and thus the minimum value is $2^n$.  I don't know how to prove it. Please help.
21307
If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger inequality. Thanks.
57193
How to prove this for positive real numbers? $$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$  I tried AM-GM, CS inequality but all failed.
54049
For positive real numbers $a,b,c$ prove that $$ \sum{\frac{a^3}{(a+b)(a+c)}} \geq \frac{a+b+c}{4}$$  Source: high school olympiad inequality  I didn't succeed in any way, I am quite sure it will be done with multiple AG inequalities
14943
If $a, b$ are positive real numbers and $a+b = 1$, prove that :  $$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$  I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.  Thank you.