prove $\sum_{cyc}\frac{1}{a(a+b)}\ge\frac{4}{ac+bd}$

Question

Let $a,b,c,d$ be positives. Show that $$\sum_{cyc}\dfrac{1}{a(a+b)}\ge\dfrac{4}{ac+bd}$$

Use Cauchy-Schwarz inequality we have $$\sum_{cyc}\dfrac{1}{a(a+b)} \cdot\sum_{cyc}a(a+b)\ge (1+1+1+1)^2$$ it suffient $$\dfrac{16}{\sum\limits_{cyc}(a^2+ab)}\ge\dfrac{4}{ac+bd}$$ However this is wrong inequality

Answer 1

By AM-GM $$\frac{2}{\sqrt{abcd}}\geq\frac{4}{ac+bd}.$$ Thus, it remains to prove that $$\sum_{cyc}\frac{1}{a^2+ab}\geq\frac{2}{\sqrt{abcd}}.$$ Since the last inequality is homogeneous, we can assume that $abcd=1$.

Now, let $a=\frac{y}{x}$, $b=\frac{z}{y}$ and $c=\frac{t}{z}$, where $x$, $y$, $z$ and $t$ are positives.

Hence, $d=\frac{x}{t}$ and we need to prove that $$\sum_{cyc}\frac{1}{\frac{y}{x}\left(\frac{y}{x}+\frac{z}{y}\right)}\geq2$$ or $$\sum_{cyc}\frac{x^2}{y^2+xz}\geq2.$$ But by C-S we have $$\sum_{cyc}\frac{x^2}{y^2+xz}=\sum_{cyc}\frac{x^4}{x^2y^2+x^3z}\geq\frac{(x^2+y^2+z^2+t^2)^2}{\sum\limits_{cyc}(x^2y^2+x^3z)}.$$ Thus, it remains to prove that $$(x^2+y^2+z^2+t^2)^2\geq2\sum\limits_{cyc}(x^2y^2+x^3z)$$ or $$(x^2+z^2)(x-z)^2+(y^2+t^2)(y-t)^2\geq0.$$ Done!

Answer 2

$$\Leftrightarrow \left ( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right )^2\cdot \dfrac{1}{4+\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}}\ge \dfrac{4}{ac+bd}$$

$$\left ( \left (\dfrac{1}{a}+\dfrac{1}{b} \right )+\left (\dfrac{1}{c}+\dfrac{1}{d}\right )\right )^2 \cdot (ac)\ge 4 \left( \dfrac{c}{b}+\dfrac{d}{a}\right)+\dfrac{2(a+b)(c+d)}{bd}$$

$$\left ( \left (\dfrac{1}{a}+\dfrac{1}{d} \right )+\left (\dfrac{1}{c}+\dfrac{1}{b}\right )\right )^2 \cdot (bd)\ge 4 \left( \dfrac{b}{a}+\dfrac{d}{c}\right)+\dfrac{2(d+a)(b+c)}{ac}$$

So:

$$ \left (\dfrac{1}{a}+\dfrac{1}{b} +\dfrac{1}{c}+\dfrac{1}{d}\right )^2\cdot(ac+bd)\ge 4\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}\right)+16$$