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Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$

Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2}, z = \dfrac{a + b}{2}$

It needs to be sufficient to prove that $$\sum_{cyc}\frac{\dfrac{a + b}{2} + \dfrac{b + c}{2} - \dfrac{c + a}{2}}{\left(2 \cdot \dfrac{c + a}{2}\right)^2} \ge \frac{9}{\displaystyle 4 \cdot \sum_{cyc}\dfrac{c + a}{2}} \implies \sum_{cyc}\frac{z + x - y}{y^2} \ge \frac{9}{y + z + x}$$

According to the Cauchy-Schwarz inequality, we have that

$$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\sqrt{\frac{z + x - y}{y}}\right)^2$$

We need to prove that $$\sum_{cyc}\sqrt{\frac{z + x - y}{y}} \ge 3$$

but I don't know how to.

Thanks to Isaac YIU Math Studio, we additionally have that $$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} = \sum_{cyc}(z + x - y) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\frac{z + x - y}{y}\right)^2$$

We now need to prove that $$\sum_{cyc}\frac{z + x - y}{y} \ge 3$$, which could be followed from Nesbitt's inequality.

I would be greatly appreciated if there are any other solutions than this one.

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  • $\begingroup$ I have a suggestion that when you do CS, try to use this: $$x+y+z=(x+y-z)+(y+z-x)+(z+x-y)$$ $\endgroup$
    – MafPrivate
    Jan 14, 2020 at 15:05
  • $\begingroup$ When I was typing out the problem, I have considered the possibility of solving the problem by letting $x = b + c - a, y = c + a - b, z = a + b - c$, but then that idea was scrapped. $\endgroup$ Jan 14, 2020 at 15:07
  • $\begingroup$ Also after multiplying with all the denominators it's equivalent to $$2\sum_{\text{sym}} a^6+12\sum_{\text{sym}} a^5b+3\sum_{\text{sym}}a^4b^2+7\sum_{\text{sym}} a^4bc\geq 5\sum_{\text{sym}} a^3b^3+14 \sum_{\text{sym}} a^3b^2c+5\sum_{\text{sym}} a^2b^2c^2$$ which is true by Muirhead $\endgroup$ Jan 14, 2020 at 16:28

8 Answers 8

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Hint : Using homogeneity, WLOG we may set $a+b+c=3$, then note $f(x) = \dfrac{x}{(3-x)^2}$ is convex and use Jensen’s inequality.

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By Cauchy-Schwarz, $$\left[\dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2}\right](a+b+c) \ge \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2$$ Then by rearrangement inequality, $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{a}{a+b} \\ \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{c}{b+c}+\dfrac{a}{c+a}+\dfrac{b}{a+b}$$ Sum up and get $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}$$ So from the first inequality, we get: $$\left[\dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2}\right](a+b+c) \ge \dfrac{9}{4} \\ \dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2} \ge \dfrac{9}{4(a+b+c)}$$

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Recall Nesbitt's inequality \begin{eqnarray*} \frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b} \geq \frac{3}{2}. \end{eqnarray*} Using Cauchy-Schwartz and Nesbitt gives \begin{eqnarray*} &\left( 2 \cdot \frac{a}{b+c}+1 \right)^2 +\left( 2 \cdot \frac{b}{c+a}+1 \right)^2 + \left(2 \cdot \frac{c}{a+b} +1 \right)^2\\ &\geq \frac{1}{3}\left( 2 \cdot \frac{a}{b+c}+2 \cdot \frac{b}{c+a} + 2 \cdot \frac{c}{a+b} +3 \right)^2 \geq 12. \end{eqnarray*} And this can be rearranged to give the inequality.

Edit: In light of Issac's answer ... By Cauchy-Schwartz, \begin{eqnarray*} \left(\dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2}\right)(a+b+c) \geq \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2 \end{eqnarray*} and the result now follows by Nesbitt's inequality.

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Since our inequality is homogeneous, we can assume that $a+b+c=3.$

Thus, $$\sum_{cyc}\frac{a}{(b+c)^2}-\frac{9}{4(a+b+c)}=\sum_{cyc}\frac{a}{(3-a)^2}-\frac{3}{4}=\sum_{cyc}\left(\frac{a}{(3-a)^2}-\frac{1}{4}\right)=$$ $$=\frac{1}{4}\sum_{cyc}\frac{-a^2+10a-9}{(3-a)^2}=\frac{1}{4}\sum_{cyc}\frac{(a-1)(9-a)}{(3-a)^2}=$$ $$=\frac{1}{4}\sum_{cyc}\left(\frac{(a-1)(9-a)}{(3-a)^2}-2(a-1)\right)=\frac{1}{4}\sum_{cyc}\frac{(a-1)^2(9-2a)}{(3-a)^2}\geq0.$$

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Let $a+b+c=p$, then consider a function $f(a)=\dfrac{a}{(p-a)^2}$, then $f''(a)=\dfrac{2(a+p)}{(p-a)^4} >0$. So from the Jensen's in equality it follows that $$\frac{f(a)+f(b)+f(c)}{3} \ge f\left[\frac{a+b+c}{3}\right]$$ So we get $$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+c)^2} \ge 3*\frac{p/3}{(p-p/3)^2}=\frac{9}{4(a+b+c)}$$

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By Hölder's inequality: $$ ..... \geq \frac{(a+b+c)^3}{4(ab+bc+ac)^2} \geq \frac{9(a+b+c)^3}{4(a+b+c)^4}$$

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Also, SOS helps: $$\sum_{cyc}\frac{a}{(b+c)^2}-\frac{9}{4(a+b+c)}=\sum_{cyc}\left(\frac{a}{(b+c)^2}-\frac{3}{4(a+b+c)}\right)=$$ $$=\tfrac{1}{4(a+b+c)}\sum_{cyc}\tfrac{4a^2+4a(b+c)-3(b+c)^2}{(b+c)^2}=\tfrac{1}{4(a+b+c)}\sum_{cyc}\tfrac{(a-b-(c-a))(2a+3b+3c)}{(b+c)^2}=$$ $$=\frac{1}{4(a+b+c)}\sum_{cyc}(a-b)\left(\frac{2a+3b+3c}{(b+c)^2}-\frac{2b+3a+3c}{(a+c)^2}\right)=$$ $$=\frac{1}{4(a+b+c)}\sum_{cyc}\frac{(a-b)^2(2a^2+2b^2+5c^2+5ab+7ac+7bc)}{(a+c)^2(b+c)^2}\geq0.$$

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Another way.

By Holder $$\sum_{cyc}\frac{a}{(b+c)^2}=\sum_{cyc}\frac{a^3}{a^2(b+c)^2}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}a^2(b+c)^2}=\frac{(a+b+c)^3}{6\sum\limits_{cyc}(a^2b^2+a^2bc)}$$ and it's enough to prove that: $$2(a+b+c)^4\geq27\sum\limits_{cyc}(a^2b^2+a^2bc)$$ or $$\sum_{cyc}(2a^4+8a^3b+8a^3c-15a^2b^2-3a^2bc)\geq0,$$ which is true by Muirhead.

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