a mathematical problem on inequalities

If $a,b,c,d$ are positive real numbers such that $a+b+c+d=1$,show that

$$ \frac{a^3}{b + c} + \frac{b^3}{c + d} + \frac{c^3}{d + a} + \frac{d^3}{a + b} > \frac 1 8$$

Solution 1:

By Cauchy-Schwarz Inequality, we have$$\left(\sum_{cyc} \frac{a^3}{b + c} \right)\cdot \sum_{cyc}(b+c) \ge (a^{3/2}+b^{3/2}+c^{3/2}+d^{3/2})^2$$

So it is sufficient to show that $a^{3/2}+b^{3/2}+c^{3/2}+d^{3/2} \ge \frac12$.

But we have from Power Means inequality, $$\left(\frac{a^{3/2}+b^{3/2}+c^{3/2}+d^{3/2}}{4}\right)^{2/3}\ge \frac{a+b+c+d}{4}=\frac14 \implies a^{3/2}+b^{3/2}+c^{3/2}+d^{3/2} \ge \frac12$$

Equality is attained when $a=b=c=d=\frac14$.