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Algebra Difficulty 7.1 National olympiad, round 2 Prove it

Example 7 Let a,b,ca, b, c be the lengths of the sides of a triangle, and s,R,rs, R, r be its semi-perimeter, circumradius, and inradius, respectively, then
(∑s−a)2≤(4+rR)∑bc∑a\left(\sum \sqrt{s-a}\right)^{2} \leq\left(4+\frac{r}{R}\right) \frac{\sum b c}{\sum a}

Solution

Prove that by letting s−a=x,s−b=y,s−c=z\sqrt{s-a}=x, \sqrt{s-b}=y, \sqrt{s-c}=z, we obtain the equivalent form of (3.2.12):
L7(x,y,z)=f0,3(8)−2f0,4(8)+2f0,5(8)+2f1,1(8)−2f1,2(8)−2f1,3(8)+4f2,1(8)≥0L_{7}(x, y, z)=f_{0,3}^{(8)}-2 f_{0,4}^{(8)}+2 f_{0,5}^{(8)}+2 f_{1,1}^{(8)}-2 f_{1,2}^{(8)}-2 f_{1,3}^{(8)}+4 f_{2,1}^{(8)} \geq 0

By combining the terms of the above expression, we get
L7(x,y,z)=(f0,3(8)−2f0,4(8))+2(f0,5(8)+f2,1(8)−f1,3(8))+2(f1,1(8)+f2,1(8)−f1,2(8))f0,3(8)−2f0,4(8)=(f0,1(2)+σ2)f0,3(6)≥0f0,5(8)+f2,1(8)−f1,3(8)=∑y2z2(x−y)2(x−z)2≥0f1,1(8)+f2,1(8)−f1,2(8)=σ3∑x(x−y)2(x−z)2≥0\begin{array}{l} L_{7}(x, y, z)=\left(f_{0,3}^{(8)}-2 f_{0,4}^{(8)}\right)+2\left(f_{0,5}^{(8)}+f_{2,1}^{(8)}-f_{1,3}^{(8)}\right)+2\left(f_{1,1}^{(8)}+f_{2,1}^{(8)}-f_{1,2}^{(8)}\right) \\ f_{0,3}^{(8)}-2 f_{0,4}^{(8)}=\left(f_{0,1}^{(2)}+\sigma_{2}\right) f_{0,3}^{(6)} \geq 0 \\ f_{0,5}^{(8)}+f_{2,1}^{(8)}-f_{1,3}^{(8)}=\sum y^{2} z^{2}(x-y)^{2}(x-z)^{2} \geq 0 \\ f_{1,1}^{(8)}+f_{2,1}^{(8)}-f_{1,2}^{(8)}=\sigma_{3} \sum x(x-y)^{2}(x-z)^{2} \geq 0 \end{array}

From the above, we can see that (3.2.12) holds, hence (3.2.12) is established. Q.E.D.

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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.