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Algebra Difficulty 6.4 National olympiad Prove it

Example 1.2.2 Let a,b,c,d>0,a+b+c+d=4a, b, c, d>0, a+b+c+d=4, prove: a1+b2c+b1+c2d+c1+d2a+d1+a2b≥2\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq 2

Solution

Prove: According to the AM-GM inequality, we have
a1+b2c=a−ab2c1+b2c≥a−ab2c2bc=a−abc2=a−ba⋅ac2≥a−b(a+ac)4\frac{a}{1+b^{2} c}=a-\frac{a b^{2} c}{1+b^{2} c} \geq a-\frac{a b^{2} c}{2 b \sqrt{c}}=a-\frac{a b \sqrt{c}}{2}=a-\frac{b \sqrt{a \cdot a c}}{2} \geq a-\frac{b(a+a c)}{4}

Based on this estimate, we have
∑cyca1+b2c≥∑cyca−14∑cycab−14∑cycabc\sum_{c y c} \frac{a}{1+b^{2} c} \geq \sum_{c y c} a-\frac{1}{4} \sum_{c y c} a b-\frac{1}{4} \sum_{c y c} a b c

Using the AM-GM inequality again, it is easy to get
∑cycab≤14(∑cyca)2=4;∑cycabc≤114(∑cyca)3=4\sum_{c y c} a b \leq \frac{1}{4}\left(\sum_{c y c} a\right)^{2}=4 ; \quad \sum_{c y c} a b c \leq \frac{1}{14}\left(\sum_{c y c} a\right)^{3}=4

Therefore
a1+b2c+b1+c2d+c1+d2a+d1+a2b≥a+b+c+d−2=2\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq a+b+c+d-2=2

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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.