Apply Cauchy-Schwarz?

Relevant wiki: Classical Inequalities - Problem Solving - Intermediate

We will first prove Titu's lemma:

Titu's Lemma : $\frac {a^2}{x} + \frac {b^2}{y} \geq \frac {(a+b)^2}{x+y}$

Multiplying denominators leads to

$\begin{aligned} a^2 y (x+y) + b^2 x (x+y) &\geq (a+b)^2 xy\\ a^2 xy + a^2 y^2 + b^2 x^2 + b^2 xy &\geq a^2 xy + b^2 xy + 2abxy\\ a^2 y^2 + b^2 x^2 - 2abxy &\geq 0\\ (ay-bx)^2 &\geq 0 \end{aligned}$

The final statement is trivially true, so our lemma must also be true. Thus, proven. Note that we can generalise this to $\displaystyle \sum \dfrac {a_i^2}{b_i} \geq \dfrac{(\displaystyle \sum a_i)^2}{\displaystyle \sum b_i}$ . The proof for this is not too hard and is left as exercise to the reader.

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Note that $\dfrac {a}{b(b+c)^2} = \dfrac {\frac {a^2}{(b+c)^2}}{ab}$ . Substituting this into the original expression for each of the other two parts as well, then applying the lemma, we yield

$\dfrac {a}{b(b+c)^2} + \dfrac {b}{c(c+a)^2} + \dfrac {c}{a(a+b)^2} \geq \dfrac{\left ( \frac {a}{b+c} + \frac {b}{c+a} + \frac {c}{a+b} \right )^2}{ab+bc+ca} = \dfrac{\left ( \frac {a}{b+c} + \frac {b}{c+a} + \frac {c}{a+b} \right )^2}{k}$

We will now prove $\frac {a}{b+c} + \frac {b}{c+a} + \frac {c}{a+b} \geq \frac {3}{2}$ . Note the following:

$\begin{aligned} \dfrac {a}{b+c} + \dfrac {b}{c+a} + \dfrac {c}{a+b} &= \dfrac {a^2}{ab+ac} + \dfrac {b^2}{bc+ab} + \dfrac {c^2}{ac+bc}\\ &\geq \dfrac {(a+b+c)^2}{2(ab+bc+ca)} \end{aligned}$

It suffices to prove $\frac {(a+b+c)^2}{2(ab+bc+ca)} \geq \frac {3}{2}$ , which simplifies to $a^2+b^2+c^2 \geq ab+bc+ca$ , which can be rearranged to $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$ , which is trivially true. Thus, $\frac {a}{b+c} + \frac {b}{c+a} + \frac {c}{a+b} \geq \frac {3}{2}$ , so $\left ( \frac {a}{b+c} + \frac {b}{c+a} + \frac {c}{a+b} \right )^2 = \frac {9}{4}$ . Thus,

$\dfrac {a}{b(b+c)^2} + \dfrac {b}{c(c+a)^2} + \dfrac {c}{a(a+b)^2} \geq \dfrac {9}{4k} = \dfrac {9}{4} k^{-1}$

so $x=9$ , $y=4$ and $z=-1$ . Thus, $x+y+z=12$ .

By the Cauchy-Schwarz inequality, $\left(\left(\sqrt{\frac{a}{b(b+c)^2}}\right)^2+\left(\sqrt{\frac{b}{c(c+a)^2}}\right)^2+\left(\sqrt{\frac{c}{a(a+b)^2}}\right)^2\right)\left((\sqrt{ab})^2+(\sqrt{bc})^2+(\sqrt{ca})^2\right)\geq \left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2$ Without loss of generality,let $a\geq b\geq c$ .Then $\dfrac{1}{b+c}\geq \dfrac{1}{a+c}\geq \dfrac{1}{a+b}$ .

Applying the Rearrangement Inequality to the sequences $(a,b,c)$ and $(\dfrac{1}{b+c},\dfrac{1}{a+c},\dfrac{1}{a+b})$ ,we get: $\begin{aligned} \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} &\geq \dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{a}{a+b}\rightarrow (1)\\ \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}&\geq \dfrac{c}{b+c}+\dfrac{a}{c+a}+\dfrac{b}{a+b}\rightarrow (2) \end{aligned}\\ (1)+(2)\implies 2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}\dfrac{c}{a+b}\right)\geq \dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}+\dfrac{a+b}{a+b}=3\\ \implies \boxed{\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq \dfrac{3}{2}}$ Hence $\dfrac{a}{b(b+c)^2}+\dfrac{b}{c(c+a)^2}+\dfrac{c}{a(a+b)^2}\geq \left(\dfrac{3}{2}\right)^2\times k^{-1}=\dfrac{9}{4}k^{-1}$ .Therefore $x=9,y=4,z=-1\implies x+y+z=\boxed{12}$