Problem
ALG-B2-M07-P020 Choose the normalization
#20
★★★★★ Level 5 of 5
Let \(a,b,c>0\). Prove \[\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{ab+bc+ca}{\sqrt[3]{a^2b^2c^2}}\ge4.\]
Hint. Normalize \(abc=1\), then use \(a^2+b^2+c^2\ge ab+bc+ca\) and \(ab+bc+ca\ge3\).
The expression has degree \(0\), so normalize \(abc=1\). Then \(\sqrt[3]{a^2b^2c^2}=1\). We get \[\frac{\sum a^2}{q}+q.\] Since \(\sum a^2\ge q\), the first fraction is at least \(1\). Also \(q=ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}=3\). Hence the whole sum is at least \(1+3=4\).
Final problem: the student must choose the normalization and combine two basic estimates.