Maths Olympiad Prep

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Algebra Difficulty 6.9 National olympiad Prove it

Example 4 Let a,b,c⩾0a, b, c \geqslant 0, and ab+bc+ca=1ab + bc + ca = 1. Prove: a31+b2+b31+c2+c31+a2⩾34\frac{a^{3}}{1+b^{2}}+\frac{b^{3}}{1+c^{2}}+\frac{c^{3}}{1+a^{2}} \geqslant \frac{\sqrt{3}}{4}.

Solution

Proof: Notice that
a31+b2=a3ab+bc+ca+b2=a3(b+a)(b+c)=[a3(b+a)(b+c)+b+a8+b+c8]−b4−a8−c8\begin{array}{l} \frac{a^{3}}{1+b^{2}}=\frac{a^{3}}{a b+b c+c a+b^{2}}=\frac{a^{3}}{(b+a)(b+c)} \\ =\left[\frac{a^{3}}{(b+a)(b+c)}+\frac{b+a}{8}+\frac{b+c}{8}\right]-\frac{b}{4}-\frac{a}{8}-\frac{c}{8} \end{array}

equals a3643−b4−a8−c8=5a8−b4−c8\sqrt[3]{\frac{a^{3}}{64}}-\frac{b}{4}-\frac{a}{8}-\frac{c}{8}=\frac{5 a}{8}-\frac{b}{4}-\frac{c}{8}.
Thus, the left side of equation (1)
=∑a31+b2⩾∑(5a8−b4−c8)=58∑a−14∑b−18∑c=58∑a−14∑a−18∑a=14∑a=14(a+b+c)2⩾143(ab+bc+ca)=34.\begin{array}{l} =\sum \frac{a^{3}}{1+b^{2}} \geqslant \sum\left(\frac{5 a}{8}-\frac{b}{4}-\frac{c}{8}\right) \\ =\frac{5}{8} \sum a-\frac{1}{4} \sum b-\frac{1}{8} \sum c \\ =\frac{5}{8} \sum a-\frac{1}{4} \sum a-\frac{1}{8} \sum a \\ =\frac{1}{4} \sum a=\frac{1}{4} \sqrt{(a+b+c)^{2}} \\ \geqslant \frac{1}{4} \sqrt{3(a b+b c+c a)}=\frac{\sqrt{3}}{4} . \end{array}

Therefore, the original inequality is proved.

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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.