Cyclic

The expression is equivalent to :

$\dfrac{a^4}{ab + ac} + \dfrac{b^4}{bc + bd} + \dfrac{c^4}{cd + ca} + \dfrac{d^4}{da + db}$ , Call it $S$

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By Titu's Lemma ,

$S \geq \dfrac{(a^2 + b^2 + c^2 + d^2 )^{2}}{ab + ac + ad + bc + bd + cd + ac + bd}$ .

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Consider the numerator ,

By C - S Inequality , $(a^ 2 + b^2 + c^2 + d^2)(1 + 1 + 1 + 1) \geq (a + b + c + d)^{2} = 1$

Therefore , $(a^2 + b^ 2 + c^2 + d^2)^{2} \geq \dfrac{1}{16} ......................... (1)$

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Now , consider the denominator ,

$ab + ac + ad + bc + bd + cd + ac + bd = (a + d)(b + c) + (a + b)(c + d)$

By AM -GM Inequality on terms $(a + d) , (b + c)$

$(a + d)(b + c) \leq (\dfrac{(a + d) + (b + c)}{2})^{2} = \dfrac{1}{4}$

Similarly $(a + b)(d + c) \leq (\dfrac{(a + b) + (d + c)}{2})^{2} = \dfrac{1}{4}$ .

Therefore, $\dfrac{1}{(a + d)(b + c) + (a + b)(c + d)} = \dfrac{1}{ab + ac + ad + bc + bd + cd + ac + bd} \geq 2$ . $....................(2)$

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From $(1) \text{and} (2)$ ,

$\dfrac{(a^2 + b^2 + c^2 + d^2)^{2}}{ab +ac + ad + bc + bd + cd + ac + bd} \geq \dfrac{1}{8}$ .

$\dfrac{a^3}{b + c} + \dfrac{b^3}{c + d} + \dfrac{c^3}{d + a} + \dfrac{d^3}{a + b} \geq \dfrac{1}{8}$ .

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Equality hold when $a = b = c = d = \dfrac{1}{4}$

By Hölder's inequality, $\left(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+d}+\dfrac{c^3}{d+a}+\dfrac{d^3}{a+b}\right)[(b+c)+(c+d)+(d+a)+(a+b)](1+1+1+1)\ge(a+b+c+d)^3$

$\Longleftrightarrow\left(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+d}+\dfrac{c^3}{d+a}+\dfrac{d^3}{a+b}\right)(2)(4)\ge1$

$\Longleftrightarrow\dfrac{a^3}{b+c}+\dfrac{b^3}{c+d}+\dfrac{c^3}{d+a}+\dfrac{d^3}{a+b}\ge\dfrac{1}{8}$

Equality holds if and only if $a=b=c=d=\dfrac{1}{4}$