Problem
ALG-B2-M03-P016 A fraction with a cube
#16
★★★★★ Level 5 of 5
Let \(a,b,c,d>0\), \(a+b+c+d=10\). Prove \[\sum_{\mathrm{cyc}}\frac{a^3}{a^2+b+c}\ge 5-\frac12(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}).\]
Hint 1. Transform one fraction: \(\frac{a^3}{a^2+b+c}=a-\frac{a(b+c)}{a^2+b+c}\).
Hint 2. Bound the denominator below by \(2a\sqrt{b+c}\).
For each term, \[\frac{a^3}{a^2+b+c}=a-\frac{a(b+c)}{a^2+b+c}\ge a-\frac{a(b+c)}{2a\sqrt{b+c}}=a-\frac{\sqrt{b+c}}2.\] Adding the four estimates and using \(a+b+c+d=10\) gives the result.
Cauchy-Schwarz module training problem. Method tags: cauchy, am-gm, bounds.