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Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$

Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rarely.

Things I have done so far: The inequality look is similar to Nesbitt's inequality.

We could re-write it as: $$\sum \limits_{cyc}\frac {a}{(b+c)^2}(2(a+b+c)) \geq \frac{9}{2}$$

Re-write it again:$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \frac{9}{2}$$ Cauchy appears: $$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2$$ So, if I prove $\left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2 \geq \frac {9}{2}$ then problem is solved.

Re-write in semi expanded form:$$2\left(\sum \limits_{cyc}\frac{a}{b+c}+2\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)}\right) \geq 9$$

We know that $\sum \limits_{cyc}\frac{a}{b+c} \geq \frac {3}{2}$.So$$4\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq 6$$

So the problem simplifies to proving this $$\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq \frac{3}{2}$$

And I'm stuck here.

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Let $x=b+c,y=c+a,z=a+b$, then $a=\frac{y+z-x}{2},\dots$. Your inequality becomes $$(x+y+z)\left(\frac{y+z-x}{x^2}+\cdots\right)\geq 9.$$

Write $y+z-x=(x+y+z)-2x,\dots$ we need to show $$(x+y+z)^2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right) -2(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\geq 9.$$

Use $3\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{x^2}\right)\geq \left(\frac1x+\frac1y+\frac1z\right)^2$, we only need to show

$$\frac13(x+y+z)^2\left(\frac1x+\frac1y+\frac1z\right)^2-2(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\geq 9.$$ The last inequality is correct because $(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\geq 9$.

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Since the inequality is homogeneous, WLOG assume that $a+b+c = 1$.

Then, the inequality becomes $\dfrac{a}{(1-a)^2}+\dfrac{b}{(1-b)^2}+\dfrac{c}{(1-c)^2} \ge \dfrac{9}{4}$.

Since the function $f(x) = \dfrac{x}{(1-x)^2}$ is concave up for $x > 0$, by Jensen's Inequality, we have:

$f(a)+f(b)+f(c) \ge 3f\left(\dfrac{a+b+c}{3}\right) = 3f\left(\dfrac{1}{3}\right) = \dfrac{9}{4}$, as desired.

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Well I found a different approach for solving this problem. We could rewrite the inequality from the question as:

$$(a+b+c)\left(\sum \limits_{cyc}\frac {a}{(b+c)^2}\right) \ge \frac{9}{4}$$

By the Cauchy-Schwarz inequality:

$$(a+b+c)\left(\sum \limits_{cyc}\frac {a}{(b+c)^2}\right) \ge \left(\sum \limits_{cyc}\frac {a}{b+c}\right)^2$$

And by Nesbitt's inequality:

$$\left(\sum \limits_{cyc}\frac {a}{b+c}\right)\ge \frac{3}{2}$$

So

$$\left(\sum \limits_{cyc}\frac {a}{b+c}\right)^2 \ge \frac{9}{4}$$

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$\sum \limits_{cyc}\dfrac {a}{(b+c)^2}=\sum \limits_{cyc}\dfrac {2a^2}{2a(b+c)^2} \ge \dfrac{2(a+b+c)^2}{2a(b+c)^2+2b(a+c)^2+2c(a+b)^2}\ge \dfrac{2(a+b+c)^2}{3\times \left(\dfrac{2a+2b+2c}{3}\right)^3}=\dfrac {9}{4(a+b+c)}$

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Let $a+b+c=3$.

Hence, we need to prove that: $$\sum_{cyc}\frac{a}{(3-a)^2}\geq\frac{3}{4}$$ or $$\sum_{cyc}\left(\frac{a}{(3-a)^2}-\frac{1}{4}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-1)(9-a)}{(3-a)^2}\geq0$$ or $$\sum_{cyc}\left(\frac{(a-1)(9-a)}{(3-a)^2}-2(a-1)\right)\geq0$$ or $$\sum_{cyc}\frac{(a-1)^2(9-2a)}{(3-a)^2}\geq0$$ and since $\{a,b,c\}\subset(0,3)$, we are done!

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Without loss of generality, we assume that $a+b+c=1$. Then, \begin{align*} \frac{a}{(b+c)^2} + \frac{b}{(a+c)^2} + \frac{c}{(b+a)^2} &= \frac{a^2}{a(b+c)^2} + \frac{b^2}{b(a+c)^2} + \frac{c^2}{c(b+a)^2}\\ &\ge \left(\frac{a}{b+c} + \frac{b}{a+c}+\frac{c}{b+a} \right)^2\\ &=\left(\frac{1}{b+c} + \frac{1}{a+c}+\frac{1}{b+a} -3 \right)^2\\ &\ge \left(\frac{3}{2} -3 \right)^2\\ &=\frac{9}{4}. \end{align*}

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