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Algebra Difficulty 5.9 AIME, harder Prove it

Let a,b,ca, b, c be three strictly positive real numbers. Furthermore, assume that abc=1a b c=1. Show that

a2b+c+b2c+a+c2a+b⩾32 \frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{3}{2}

Solution

This is a particular case of the "bad students' inequality":

a2b+c+b2c+a+c2a+b⩾(a+b+c)22(a+b+c)(CS)3abc32=32 \begin{aligned} \frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} & \geqslant \frac{(a+b+c)^{2}}{2(a+b+c)} \\ & \stackrel{(\text{CS})}{ } \frac{3 \sqrt[3]{a b c}}{2}=\frac{3}{2} \end{aligned}

Here is a generalization of the Cauchy-Schwarz inequality (to solve this exercise, we can assume p p and q q are rational):

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.