I'll prove: $\sum\frac{1}{2c^{2}+2ac+2bc+3ab}\leq\frac{1}{\sum ab}$ $\leftrightarrow \sum\frac{2c(a+b-2c)}{2c^{2}+2ac+2bc+3ab}\leq 0$ $\leftrightarrow \sum\frac{a+b-2c}{2\sum a+3\frac{ab}{c}}\leq 0$(*) Finally,use chebyshev's inequality to prove (*)

silver1989 wrote: I'll prove: $\sum\frac{1}{2c^{2}+2ac+2bc+3ab}\leq\frac{1}{\sum ab}$ Maybe $\sum\frac{1}{4c^{2}+2ac+2bc+ab}\leq\frac{1}{\sum ab}$ $?$

Observe that $\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}\geq \frac{1}{ab+bc+ac}$ is equivalent to (algebraic manipulations only, us condition $a+b+c=1$ to homogenize firstly) $2 \geq \frac{1}{2+\frac{c}{a}}+\frac{1}{2+\frac{a}{c}}+\frac{1}{2+\frac{b}{a}}+\frac{1}{2+\frac{a}{b}}+\frac{1}{2+\frac{b}{c}}+\frac{1}{2+\frac{c}{b}}$ which is easy

Very nice proof, Megus!

Maybe $\sum_{cyc}\frac{1}{(2a+b)(2a+c)}\geq\sqrt{\frac{a^{2}+b^{2}+c^{2}}{(a^{2}+2bc)(b^{2}+2ac)(c^{2}+2ab)}}$ still true?

Nice Problem! My solution: see that $\frac{ab+bc+ca}{(2a+b)(2a+c)}= \frac{b(2a+c)+c(2a+b)}{2(2a+b)(2a+c)}$ $= \frac{b}{2(2a+b)}+\frac{c}{2(2a+c)}$ now, do in this same way with the other fractions. let S our sum! by cauchy, we get: $(b(2a+b)+c(2a+c)+a(2b+a)+c(2b+c)+a(2c+a)+b(2c+b)) . S$ $\geq (\frac{2a+2b+2c}{\sqrt{2}})^{2}$ $\Longrightarrow S \geq 1$

Megus wrote: $2 \geq \frac{1}{2+\frac{c}{a}}+\frac{1}{2+\frac{a}{c}}+\frac{1}{2+\frac{b}{a}}+\frac{1}{2+\frac{a}{b}}+\frac{1}{2+\frac{b}{c}}+\frac{1}{2+\frac{c}{b}}$ How this come?

dragonfire wrote: Megus wrote: $2 \geq \frac{1}{2+\frac{c}{a}}+\frac{1}{2+\frac{a}{c}}+\frac{1}{2+\frac{b}{a}}+\frac{1}{2+\frac{a}{b}}+\frac{1}{2+\frac{b}{c}}+\frac{1}{2+\frac{c}{b}}$ How this come? $\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}\geq \frac{1}{ab+bc+ac}\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{ab+ac+bc}{ab+2c^{2}+2c(a+b+c)}\geq1\Leftrightarrow2\geq\sum_{cyc}\left(1-\frac{ab+ac+bc}{4c^{2}+2ac+2bc+ab}\right)\Leftrightarrow$ $\Leftrightarrow2\geq\sum_{cyc}\frac{c(2c+a)+c(2c+b)}{(2c+a)(2c+b)}\Leftrightarrow2\geq\sum_{cyc}\left(\frac{c}{2c+b}+\frac{c}{2c+a}\right).$

$\frac{ab}{ab+ac+bc}\leq\frac{ab+ac+bc}{ab+2c^{2}+2c}$ (which is easy to show) and the rest is easy.

Megus wrote: Observe that $\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}\geq \frac{1}{ab+bc+ac}$ is equivalent to (algebraic manipulations only, us condition $a+b+c=1$ to homogenize firstly) $2 \geq \frac{1}{2+\frac{c}{a}}+\frac{1}{2+\frac{a}{c}}+\frac{1}{2+\frac{b}{a}}+\frac{1}{2+\frac{a}{b}}+\frac{1}{2+\frac{b}{c}}+\frac{1}{2+\frac{c}{b}}$ which is easy NOTE Observe now that $\frac{1}{2+x}+\frac{1}{2+\frac{1}{x}}\leq \frac{2}{3}$ Use this for $x = \frac{a}{b}, x = \frac{b}{c},x = \frac{c}{a}$ and we have a simple solution , as Megus pointed out , too ! Babis

Or (more general form) consider the function $f(x)=\frac{1}{2+x}\iff f(x-2)=\frac{1}{x}$ which is clearly concave and thus by Jensen: \[\sum^{n}\frac{1}{2+x_{i}}\leq \frac{n}{2+\frac{\sum^{n}x_{1}}{n}}\leq \frac{n}{2+\prod^{n}\sqrt[n]{x_{i}}}\] where the last is by Am-Gm. Our ineq. is the case of $n=6, \prod x_{i}=1$ when \[\frac{n}{2+\prod^{n}\sqrt[n]{x_{i}}}=\frac{6}{2+1}=2\]

me@home wrote: Or (more general form) consider the function $f(x)=\frac{1}{2+x}\iff f(x-2)=\frac{1}{x}$ which is clearly concave ... Sorry , but is this funtion concave ? I see it is convex on $(0,+\infty)$. ( I find $f''(x) =+\frac{2}{(x+2)^{3}}$). But I'll try it again ! Babis

Quote: Let $ a, b, c$ be positive reals such that their sum is $ 1$. Prove that \[ \frac {1}{ab + 2c^{2} + 2c} + \frac {1}{bc + 2a^{2} + 2a} + \frac {1}{ac + 2b^{2} + 2b}\geq \frac {1}{ab + bc + ac}.\] By $ AM-GM$, we obtain: $ a^2c^2+b^2c^2\ge 2abc^2$. Which is equivalent to: $ a^2b^2+b^2c^2+c^2a^2+2abc\ge a^2b^2+2abc^2+2abc$ or $ a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)\ge ab(ab+2c^2+2c)$ $ \Leftrightarrow$ $ (ab+ca+bc)^2\ge ab(ab+2c^2+2c)$ $ \Leftrightarrow$ $ \frac {1}{ab+2c^2+2c}\ge \frac {ab}{(ab+bc+ca)^2}$. Having the same argument, we conclude that: $ LHS$ $ \ge \frac {ab+bc+ca}{(ab+ca+bc)^2}=\frac {1}{ab+bc+ca}$ $ =RHS$. $ \square$ Our proof is completed then.

This problem has actually been posted before. See here.

$ \sum_{cyc}\frac {1}{2c^{2} + 2c + ab}\geq\frac {1}{ab + bc + ac}$ $ 2c^{2} + 2c + ab = 2c^{2} + 2(a + b + c)c + ab = (2c + a)(2c + b)$ $ \implies$ $ \sum_{cyc}\frac {2}{(2c + a)(2c + b)}\geq \frac {2}{ab + bc + ac}$ $ \implies$ $ \sum_{cyc}\frac {2(ab + ac + bc)}{(2c + a)(2c + b)}\geq 2$ $ 2(ab + ac + bc) = b(2c + a) + a(2c + b)$ $ \implies$ $ \sum_{cyc}\frac {2(ab + ac + bc)}{(2c + a)(2c + b)} = \sum_{cyc}(\frac {b}{2c + b} + \frac {a}{2c + a})$ $ \implies$ $ \sum_{cyc}(\frac {b}{2c + b} + \frac {a}{2c + a})(\sum_{cyc}{b(2c + b) + a(2c + a))\geq 4(a + b + c)^{2}}$ $ (\sum_{cyc}{b(2c + b) + a(2c + a))} = 2(a + b + c)^{2}$ $ \implies$ $ \sum_{cyc}\frac {2(ab + ac + bc)}{(2c + a)(2c + b)}\geq 2$ Q.E.D Secondly we can use muirhead inequality,but it is very long solution

Using $uvw$ method write the terms in a function with $w^3$ and fix u and v. Suppose that the function will be like that \[\sum\frac{1}{4c^{2}+2ac+2bc+ab}\ge\frac{1}{\sum ab} \Longleftrightarrow g(w^3)=Aw^6+Bw^3+C\ge0\] it is easy to see $A$ is negative so $g$ will be concave hence we must look for the minimum and the max of $w^3$ by $uvw$ theorem we must look the cases $a=b$ and $c=0$ which is obvius.

Consider the lagrange function $f(a,b,c,\lambda)=\sum\frac {1}{ab+2c^2+2c}+\frac {1}{\sum ab}-\lambda(\sum a)$ Now $f'(a)=-\sum\frac {b}{(ab+2c^2+2c)^2}-\frac {3a+2}{(bc+2a^2+2a)^2}+\frac {b+c}{(ab+bc+ac)^2}-\lambda a$. So basically now we've like $\lambda(a+t)+\frac {a^2(3a+2)}{(m+2a^3+2a^2)^2}=g$ where $t,g,m$ are constant. So $a>b>c$ implies $g(x)$ is decreasing , where $g(x)=\frac {x^2(3x+2)}{(m+2x^3+2x^2)^2}$. Now $g'(x)<0$ gives $(2a^3+2a)(9a^2+4)<(m+2a^3+2a)(9a^2+4)<2a^2(3a+2)(6a^2+2)$ now that implies $a^2+a>\frac {2}{3}\implies a>\frac {1}{2}$. Now as $g$ is decreasing in the interval $(\frac {1}{2},\inf )$ so by the condition $a,b,c\in\(\frac {1}{2},\inf )$ and that implies $\sum a>1$, absurd, similarly we can do this for if two of the are equal, now so rest and last case is we must have $a=b=c$ to obtain minimum value and that is indeed zero.

Excuse me for opening this old topic, but this inequality has another short proof that I found and want to share. Using Am-gm we have: $(2a+b)(2a+c)=\frac{1}{bc}(2ac+bc)(2ab+bc)\le \frac{1}{bc}(ab+ac+bc)^2$ Therefore: $\sum \frac{1}{ab+2c^2+2c}=\sum \frac{1}{(2a+b)(2a+c)}\ge \sum \frac{ab}{(ab+ac+bc)^2} =\frac{1}{ab+ac+bc}$ . $\blacksquare$

Here is an elementary proof ..hope that I haven't made any mistake First bring the inequality in the form $\sum\frac{(a+b)^2}{(a+b)^{2}(4c^2+2cb+2ca +ab)}\geq \frac {1}{\sum ab}$ From Cauchy-Schwartz the inequality reduces to $4(\sum ab)(\sum a^2) +8(\sum ab )^2\geq \sum (a+b)^{2}(4c^2+2cb+2ca +ab) $ or equivalently $\sum a^{3}b +\sum ab^{3} \geq 2 \sum( ab)^2$ ... Which is true from AM-GM

Let $a, b, c$ be positive reals such that their sum is $1$. Prove that \[\frac{a}{bc+2a^{2}+2a}+\frac{b}{ca+2b^{2}+2b}+\frac{c}{ab+2c^{2}+2c}\geq \frac{1}{3(bc+ca+ab)}.\]http://www.artofproblemsolving.com/community/c6h610084p3626533 Let $a,b,c$ be positive real numbers. Prove that $$\frac{a}{\sqrt{(2a+b)(2a+c)}} +\frac{b}{\sqrt{(2b+c)(2b+a)}} +\frac{c}{\sqrt{(2c+a)(2c+b)}} \leq 1$$$$\leq \sqrt{\frac{bc}{(2a+b)(2a+c)}}+\sqrt{\frac{ca}{(2b+c)(2b+a)}}+\sqrt{\frac{ab}{(2c+a)(2c+b)}}\leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}} $$$$\sqrt{\frac{a^2+bc}{(2a+b)(2a+c)}}+\sqrt{\frac{b^2+ca}{(2b+c)(2b+a)}}+\sqrt{\frac{c^2+ab}{(2c+a)(2c+b)}}\ge \sqrt{2}$$

sqing wrote: Let $a, b, c$ be positive reals such that their sum is $1$. Prove that \[\frac{a}{bc+2a^{2}+2a}+\frac{b}{ca+2b^{2}+2b}+\frac{c}{ab+2c^{2}+2c}\geq \frac{1}{3(bc+ca+ab)}.\] It's easy S.O.S

Clearing the denominator, we get $4(\sum_{cyc}a^4b^2+\sum_{cyc}a^4c^2)+4(\sum_{cyc}a^4bc)+4(\sum_{cyc}a^3b^3)\geq (\sum_{cyc}a^3b^2c+\sum_{cyc}a^3bc^2)+42a^2b^2c^2$ This holds by Muirhead inequality or AM-GM inequality

Notice that $\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}\geq \frac{1}{ab+bc+ac}$ is equivalent to $2 \geq \frac{1}{2+\frac{c}{a}}+\frac{1}{2+\frac{a}{c}}+\frac{1}{2+\frac{b}{a}}+\frac{1}{2+\frac{a}{b}}+\frac{1}{2+\frac{b}{c}}+\frac{1}{2+\frac{c}{b}}$ after homogenizing. However $\frac{1}{2+x}+\frac{1}{2+\frac{1}{x}}=\frac{4+x+\frac{1}{x}}{5+2(x+\frac{1}{x})}=\frac{4+y}{5+2y}$. Yet $\frac{4+y}{5+2y}$ is montonically decreasing and $y \geq 2$, so $\frac{4+y}{5+2y} \leq \frac{2}{3}$. The result follows.

Litlle 1000t wrote: Let $a, b, c$ be positive reals such that their sum is $1$. Prove that \[\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}\geq \frac{1}{ab+bc+ac}.\] sqing wrote: Let $a, b, c$ be positive reals such that their sum is $1$. Prove that \[\frac{a}{bc+2a^{2}+2a}+\frac{b}{ca+2b^{2}+2b}+\frac{c}{ab+2c^{2}+2c}\geq \frac{1}{3(bc+ca+ab)}.\]http://www.artofproblemsolving.com/community/c6h610084p3626533 $$ 2a^{2} + 2a + bc = 2a^{2} + 2(a + b + c)a + bc = (2a + b)(2a + c)$$$$\sum \frac{1}{bc+2a^{2}+2a}=\sum \frac{1}{(2a+b)(2a+c)},\sum \frac{a}{bc+2a^{2}+2a}=\sum \frac{a}{(2a+b)(2a+c)}$$ Let $a,b,c $ be positive reals . Prove that : $$\frac{1}{ab+bc+ca}\le \frac{1}{(2a+b)(2a+c)}+\frac{1}{(2b+c)(2b+a)}+\frac{1}{(2c+a)(2c+b)}\le \frac{1}{3\sqrt[3]{a^2b^2c^2}}$$$$\frac{a}{(2a+b)(2a+c)}+\frac{b}{(2b+c)(2b+a)}+\frac{c}{(2c+a)(2c+b)}\le \frac{1}{a+b+c}$$

Let $a,b,c $ be positive reals . Prove that : $$\frac{a}{(2a+b)(2a+c)}+\frac{b}{(2b+c)(2b+a)}+\frac{c}{(2c+a)(2c+b)}\ge \frac{a+b+c}{3(a^2+b^2+c^2)}$$

sqing wrote: Let $a,b,c $ be positive reals . Prove that : $$\frac{a}{(2a+b)(2a+c)}+\frac{b}{(2b+c)(2b+a)}+\frac{c}{(2c+a)(2c+b)}\ge \frac{a+b+c}{3(a^2+b^2+c^2)}$$ $f(x)=\frac{1}{x}$ is convex. $\sum \frac{a}{a+b+c}\cdot \frac{1}{(2a+b)(2a+c)} \ge \frac{3}{4(a^2+b^2+c^2)+5(ab+bc+ca)} \ge \frac{1}{3(a^2+b^2+c^2)}$

sqing wrote: $$\frac{a}{(2a+b)(2a+c)}+\frac{b}{(2b+c)(2b+a)}+\frac{c}{(2c+a)(2c+b)}\le \frac{1}{a+b+c}$$ It is equivalent to $\sum (a-b)(a-c)\cdot \frac{1}{(2a+b)(2a+c)} \ge 0$. Wlog $a \ge b\ge c$. We have $\frac{1}{(2a+b)(2a+c)}\le \frac{1}{(2b+c)(2b+a)} \le \frac{1}{(2c+a)(2c+b)}$ which implies the result from generalization of Schur Inequality.

It is equivalent to $$\frac{2(ab+bc+ca)}{ab+2c^2+2c}+\frac{2(ab+bc+ca)}{bc+2a^2+2a}+\frac{2(ab+bc+ca)}{ac+2b^2+2b}\ge 2$$We have $$\frac{2(ab+bc+ca)}{ab+2c^2+2c}=\frac{2(ab+bc+ca)}{ab+2c^2+2c(a+b+c)}=\frac{b(2c+a)+a(2c+b)}{(2c+a)(2c+b)}=\frac b{2c+b}+\frac a{2c+a}$$Hence, we need to prove that $$\frac a{2c+a}+\frac b{2c+b}+\frac b{2a+b}+\frac c{2a+c}+\frac a{2b+a}+\frac c{2b+c}\ge 2$$Now, just see that $$\frac a{2b+a}+\frac b{2a+b}=\frac {a^2}{2ab+a^2}+\frac {b^2}{2ab+b^2}\ge \frac{(a+b)^2}{a^2+b^2+4ab}\ge \frac 23$$The rest follows.

Nice inequality to generalize

Let $a,b,c$ positive reels such that $a+b+c=\frac{2}{3}$. Then prove the following $$\frac{1}{3ab+4c^2+4c}+\frac{1}{3bc+4a^2+4a}+\frac{1}{3ca+4b^2+4b}\geq \frac{1}{3(ab+bc+ca)}$$

Let $a,b,c$ positive reels such that $a+b+c=\frac{7}{12}$. Then prove the following $$\frac{1}{6ab+7c^2+7c}+\frac{1}{6bc+7a^2+7a}+\frac{1}{6ca+7b^2+7b}\geq \frac{1}{6(ab+bc+ca)}$$

Let $a,b,c$ positive reels such that $a+b+c=\frac{5}{8}$. Then prove the following $$\frac{1}{4ab+5c^2+5c}+\frac{1}{4bc+5a^2+5a}+\frac{1}{4ca+5b^2+5b}\geq \frac{1}{4(ab+bc+ca)}$$

Let $a,b,c$ positive reels such that $a+b+c=\frac{3}{4}$. Then prove the following $$\frac{1}{2ab+3c^2+3c}+\frac{1}{2bc+3a^2+3a}+\frac{1}{2ca+3b^2+3b}\geq \frac{1}{2(ab+bc+ca)}$$

Generalization 1 Let $a,b,c$ positive reels and $k\geq 1$ such that $a+b+c=\frac{k+1}{2k}$. Then prove the following $$\frac{1}{kab+(k+1)c^2+(k+1)c}+\frac{1}{kbc+(k+1)a^2+(k+1)a}+\frac{1}{kca+(k+1)b^2+(k+1)b}\geq \frac{1}{k(ab+bc+ca)}$$

Megus wrote: Observe that $\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}\geq \frac{1}{ab+bc+ac}$ is equivalent to (algebraic manipulations only, us condition $a+b+c=1$ to homogenize firstly) $2 \geq \frac{1}{2+\frac{c}{a}}+\frac{1}{2+\frac{a}{c}}+\frac{1}{2+\frac{b}{a}}+\frac{1}{2+\frac{a}{b}}+\frac{1}{2+\frac{b}{c}}+\frac{1}{2+\frac{c}{b}}$ which is easy arqady wrote: Very nice proof, Megus! Indeed