Inequalities # 1

$\dfrac{a^3}{b+c+d}+\dfrac{b^3}{c+d+a}+\dfrac{c^3}{d+a+b}+\dfrac{d^3}{a+b+c}$

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

$\ge\dfrac{(a^2+b^2+c^2+d^2)^2}{ab+ac+ad+bc+bd+ab+cd+ac+bc+ad+bd+cd}$ (Cauchy Schwarz inequality)

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

$\ge\dfrac{(a^2+b^2+c^2+d^2)^2}{2\left(1+\sqrt{(a^2+b^2)(c^2+d^2)}\right)}$ (Cauchy Schwarz inequality)

$\ge\dfrac{(a^2+b^2+c^2+d^2)^2}{2\left(1+\frac{a^2+b^2+c^2+d^2}{2}\right)}$ (AM-GM inequality)

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

$\ge\dfrac{1}{3}$

Note that $a^2+b^2+c^2+d^2\ge ab+bc+cd+da=1$ .

Since $f(x)=\dfrac{x^2}{2+x}$ is an increasing function at $x\ge 1$ , minimum occurs when $x$ is minimize, that is $x=1$ .

Equality holds iff $a=b=c=d=\dfrac{1}{2}$ .