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Algebra Difficulty 8.0 Shortlist Prove it

143. ("Home of Math Olympiad" website, 2009.01.19 "ppppqqqq" provided) Let x,y,z∈R− x, y, z \in \mathbf{R}^{-} , and x2+y2+z2=1 x^{2}+y^{2}+z^{2}=1 , then
∑11−yz⩽4∑x∑yz(y+z)(z+x)(x+y)\sum \frac{1}{1-y z} \leqslant \frac{4 \sum x \sum y z}{(y+z)(z+x)(x+y)}

Equality holds if and only if x=y=z=33 x=y=z=\frac{\sqrt{3}}{3} , or one of x,y,z x, y, z is zero, and the other two are both equal to 22 \frac{\sqrt{2}}{2} .

Solution

143. Simplify: Homogenize the original expression to get
∑x2+y2+z2x2+y2+z2−yz⩽4∑x∑yz(y+z)(z+x)(x+y)\sum \frac{x^{2}+y^{2}+z^{2}}{x^{2}+y^{2}+z^{2}-yz} \leqslant \frac{4 \sum x \sum yz}{(y+z)(z+x)(x+y)}

Let s1=∑x=1,s2=∑yz,s3=xyzs_{1}=\sum x=1, s_{2}=\sum yz, s_{3}=xyz, and after removing the denominators and simplifying, we get
 Equation (1) ⇔(s2−8s22+20s23−16s24)+(3−17s2+38s22−32s23)s3+(1−6s2)s32⩾0⇔s2(1−4s2)(1−2s2)2+(1−2s2)(3−11s2+16s22)s3+(1−6s2)s32⩾0\begin{aligned} \text { Equation (1) } \Leftrightarrow & \left(s_{2}-8 s_{2}^{2}+20 s_{2}^{3}-16 s_{2}^{4}\right)+\left(3-17 s_{2}+38 s_{2}^{2}-32 s_{2}^{3}\right) s_{3}+\left(1-6 s_{2}\right) s_{3}^{2} \geqslant 0 \Leftrightarrow \\ & s_{2}\left(1-4 s_{2}\right)\left(1-2 s_{2}\right)^{2}+\left(1-2 s_{2}\right)\left(3-11 s_{2}+16 s_{2}^{2}\right) s_{3}+\left(1-6 s_{2}\right) s_{3}^{2} \geqslant \\ & 0 \end{aligned}

When 1−4s2⩾01-4 s_{2} \geqslant 0, and 1−6s2⩾01-6 s_{2} \geqslant 0, Equation (2) is obviously true;
When 1−4s2⩾01-4 s_{2} \geqslant 0, and 1−6s2⩽01-6 s_{2} \leqslant 0, since s22⩾3s2s3=3s3s_{2}^{2} \geqslant 3 s_{2} s_{3}=3 s_{3}, to prove Equation (2) is true, we only need to prove □\square
(1−2s2)(3−11s2+16s22)s3+13(1−6s2)s22s3⩾0\left(1-2 s_{2}\right)\left(3-11 s_{2}+16 s_{2}^{2}\right) s_{3}+\frac{1}{3}\left(1-6 s_{2}\right) s_{2}^{2} s_{3} \geqslant 0

Since
13(1−6s2)s22s3⩾−13(1−2s2)s22s3\frac{1}{3}\left(1-6 s_{2}\right) s_{2}^{2} s_{3} \geqslant-\frac{1}{3}\left(1-2 s_{2}\right) s_{2}^{2} s_{3}

Therefore, we only need to prove
(1−2s2)(3−11s2+16s22)s3−13(1−2s2)s22s3⩾0\left(1-2 s_{2}\right)\left(3-11 s_{2}+16 s_{2}^{2}\right) s_{3}-\frac{1}{3}\left(1-2 s_{2}\right) s_{2}^{2} s_{3} \geqslant 0

It is easy to prove the above inequality, so Equation (2) is true.
When 1−4s2⩽01-4 s_{2} \leqslant 0, let s2=1−w23(w⩾0)s_{2}=\frac{1-w^{2}}{3}(w \geqslant 0), according to Theorem 2 Corollary 3 in the appendix of Chapter 4 "Application of Basic Inequalities to Prove Inequalities", s3⩾(1−2w)(1+w)227s_{3} \geqslant \frac{(1-2 w)(1+w)^{2}}{27}, at this time, we have
(1−2s2)(3−11s2+16s22)s3+(1−6s2)s32−(1−2s2)(3−11s2+16s22)(1−2w)(1+w)227−(1−6s2)[(1−2w)(1+w)227]2⩾[s3−(1−2w)(1+w)227]∣ (1−2s2)(3−11s2+16s22)+(1−6s2)[s3+(1−2w)(1+w)227]}⩾[s3−(1−2w)(1+w)227][(1−2s2)(3−11s2+16s22)+2(1−6s2)s3]⩾[s3−(1−2w)(1+w)227][(1−2s2)(3−11s2+16s22)+23(1−6s2)s22]\begin{array}{l} \left(1-2 s_{2}\right)\left(3-11 s_{2}+16 s_{2}^{2}\right) s_{3}+\left(1-6 s_{2}\right) s_{3}^{2}- \\ \left(1-2 s_{2}\right)\left(3-11 s_{2}+16 s_{2}^{2}\right) \frac{(1-2 w)(1+w)^{2}}{27}- \\ \left(1-6 s_{2}\right)\left[\frac{(1-2 w)(1+w)^{2}}{27}\right]^{2} \geqslant \\ {\left.\left[s_{3}-\frac{(1-2 w)(1+w)^{2}}{27}\right] \right\rvert\,\left(1-2 s_{2}\right)\left(3-11 s_{2}+16 s_{2}^{2}\right)+} \\ \left.\left(1-6 s_{2}\right)\left[s_{3}+\frac{(1-2 w)(1+w)^{2}}{27}\right]\right\} \geqslant \\ {\left[s_{3}-\frac{(1-2 w)(1+w)^{2}}{27}\right]\left[\left(1-2 s_{2}\right)\left(3-11 s_{2}+16 s_{2}^{2}\right)+2\left(1-6 s_{2}\right) s_{3}\right] \geqslant} \\ {\left[s_{3}-\frac{(1-2 w)(1+w)^{2}}{27}\right]\left[\left(1-2 s_{2}\right)\left(3-11 s_{2}+16 s_{2}^{2}\right)+\frac{2}{3}\left(1-6 s_{2}\right) s_{2}^{2}\right]} \end{array}
(Notice that when 1−4s2⩽01-4 s_{2} \leqslant 0, we have 1−6s2⩽01-6 s_{2} \leqslant 0) we get
 The above expression ⩾0\text { The above expression } \geqslant 0

That is, □\square
(1−2s2)(3−11s2+16s22)s3+(1−6s2)s32⩾(1−2s2)(3−11s2+16s22)(1−2w)(1+w)227+(1−6s2)[(1−2w)(1+w)227]2\begin{array}{l} \left(1-2 s_{2}\right)\left(3-11 s_{2}+16 s_{2}^{2}\right) s_{3}+\left(1-6 s_{2}\right) s_{3}^{2} \geqslant \\ \left(1-2 s_{2}\right)\left(3-11 s_{2}+16 s_{2}^{2}\right) \frac{(1-2 w)(1+w)^{2}}{27}+ \\ \left(1-6 s_{2}\right)\left[\frac{(1-2 w)(1+w)^{2}}{27}\right]^{2} \end{array}

Therefore, to prove Equation (2) is true, we only need to prove
1−w23⋅[1−4(1−w2)3][1−2(1−w2)3]2+[1−2(1−w2)3][3−11(1−w2)3+16(1−w23)2]⋅(1−2w)(1+w)227+(1−6(1−w2)3)[(1−2w)(1+w)227]2⩾0\begin{array}{l} \frac{1-w^{2}}{3} \cdot\left[1-\frac{4\left(1-w^{2}\right)}{3}\right]\left[1-\frac{2\left(1-w^{2}\right)}{3}\right]^{2}+ \\ {\left[1-\frac{2\left(1-w^{2}\right)}{3}\right]\left[3-\frac{11\left(1-w^{2}\right)}{3}+16\left(\frac{1-w^{2}}{3}\right)^{2}\right] \cdot} \\ \frac{(1-2 w)(1+w)^{2}}{27}+\left(1-\frac{6\left(1-w^{2}\right)}{3}\right)\left[\frac{(1-2 w)(1+w)^{2}}{27}\right]^{2} \geqslant 0 \end{array}

After simplifying, we get □\square
(1+w)(2w−1)272(−8w2+8w3−50w4+28w5−100w6−32w7)⩾0\frac{(1+w)(2 w-1)}{27^{2}}\left(-8 w^{2}+8 w^{3}-50 w^{4}+28 w^{5}-100 w^{6}-32 w^{7}\right) \geqslant 0

Additionally, from 1−4s2⩽01-4 s_{2} \leqslant 0, we get 2w−1⩽02 w-1 \leqslant 0, and 1+w>01+w>0, so we only need to prove
−8w2+8w3−50w4+28w5−100w6−32w7⩽0-8 w^{2}+8 w^{3}-50 w^{4}+28 w^{5}-100 w^{6}-32 w^{7} \leqslant 0

In fact, we have
−8w2+8w3−50w4+28w5−100w6−32w7=−8w2(1−w2)2−14w4(1−w)2−34w4−86w6−32w7⩽0\begin{array}{l} -8 w^{2}+8 w^{3}-50 w^{4}+28 w^{5}-100 w^{6}-32 w^{7}= \\ -8 w^{2}\left(1-\frac{w}{2}\right)^{2}-14 w^{4}(1-w)^{2}-34 w^{4}-86 w^{6}-32 w^{7} \leqslant \\ 0 \end{array}

Therefore, when 1−4s2⩽01-4 s_{2} \leqslant 0, Equation (2) is also true.
In summary, Equation (1) is true, and it is easy to prove that the equality holds if and only if x=y=z=33x=y=z=\frac{\sqrt{3}}{3}, or one of x,y,zx, y, z is zero, and the other two are 22\frac{\sqrt{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.