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Algebra Difficulty 6.3 National olympiad Prove it

Example 8.3.6 Let a,b,c,d≥0a, b, c, d \geq 0, prove that
ab2+c2+d2+bc2+d2+a2+cd2+a2+b2+da2+b2+c2≥332⋅1a2+b2+c2+d2\frac{a}{b^{2}+c^{2}+d^{2}}+\frac{b}{c^{2}+d^{2}+a^{2}}+\frac{c}{d^{2}+a^{2}+b^{2}}+\frac{d}{a^{2}+b^{2}+c^{2}} \geq \frac{3 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}}

Solution

Prove: Without loss of generality, assume a2+b2+c2+d2=1a^{2}+b^{2}+c^{2}+d^{2}=1, then the inequality becomes
a1−a2+b1−b2+c1−c2+d1−d2≥332\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}}+\frac{d}{1-d^{2}} \geq \frac{3 \sqrt{3}}{2}

By the AM-GM inequality, we have
2a2(1−a2)(1−a2)≤(23)3⇒a(1−a2)≤233⇒a1−a2≥332a22 a^{2}\left(1-a^{2}\right)\left(1-a^{2}\right) \leq\left(\frac{2}{3}\right)^{3} \Rightarrow a\left(1-a^{2}\right) \leq \frac{2}{3 \sqrt{3}} \Rightarrow \frac{a}{1-a^{2}} \geq \frac{3 \sqrt{3}}{2} a^{2}

Therefore, we have
∑cyca1−a2≥332(∑cyc a2)=332\sum_{c y c} \frac{a}{1-a^{2}} \geq \frac{3 \sqrt{3}}{2}\left(\sum_{\text {cyc }} a^{2}\right)=\frac{3 \sqrt{3}}{2}

Equality holds when a=b=c,d=0a=b=c, d=0 and their permutations.

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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.