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Algebra Difficulty 7.6 National olympiad, round 2 Prove it

109. (Chen Ji, 2007. 12. 20 provided) Let x,y,z∈R−x, y, z \in \mathbf{R}^{-}, and satisfy x+y+z=2x+y+z=2, then
∑9−8yz⩾7\sum \sqrt{9-8 y z} \geqslant 7

Equality holds if and only if x=y=z=23x=y=z=\frac{2}{3}, or one of x,y,zx, y, z is zero, and the other two are equal to 1.

Solution

109. Simplified Proof: For the convenience of proof, we make the following transformation: Let x=2a,y=2b,z=2cx=2a, y=2b, z=2c. At this point, the original proposition is equivalent to:

Let a,b,c∈R−‾a, b, c \in \overline{\mathbf{R}^{-}}, and satisfy a+b+c=1a+b+c=1, then
∑9−32bc⩾7\sum \sqrt{9-32bc} \geqslant 7

Equality holds if and only if a=b=c=13a=b=c=\frac{1}{3}, or one of a,b,ca, b, c is zero, and the other two are both 12\frac{1}{2}.

Next, we will prove equation (1), i.e.,
Equation (1)⇔(9−32bc+9−32ac)2⩾(7−9−32ab)2⇔9−32bc⋅9−32ac+79−32ab⩾20+16bc+16ac−16ab\begin{aligned} \text{Equation (1)} \Leftrightarrow & (\sqrt{9-32bc}+\sqrt{9-32ac})^{2} \geqslant (7-\sqrt{9-32ab})^{2} \Leftrightarrow \\ & \sqrt{9-32bc} \cdot \sqrt{9-32ac} + 7 \sqrt{9-32ab} \geqslant \\ & 20 + 16bc + 16ac - 16ab \end{aligned}

Squaring both sides of the above equation and rearranging, we get
7729−2592∑bc+9216abc−32768(abc)2⩾−61+128(∑bc)2−512abc+464∑bc\begin{array}{l} 7 \sqrt{729 - 2592 \sum bc + 9216 abc - 32768 (abc)^{2}} \geqslant \\ -61 + 128 \left(\sum bc\right)^{2} - 512 abc + 464 \sum bc \end{array}

Let s1=∑a=1,s2=∑bc,s3=abcs_{1} = \sum a = 1, s_{2} = \sum bc, s_{3} = abc, then the above equation can be written as
7729−2592s2+9216s3−32768s32⩾−61+128s22−512s3+464s2\begin{array}{l} 7 \sqrt{729 - 2592 s_{2} + 9216 s_{3} - 32768 s_{3}^{2}} \geqslant \\ -61 + 128 s_{2}^{2} - 512 s_{3} + 464 s_{2} \end{array}

Therefore, to prove equation (1), we only need to prove equation (2) holds.
Since
s13−4s1s2+9s3⩾0s_{1}^{3} - 4 s_{1} s_{2} + 9 s_{3} \geqslant 0
(Refer to Chapter 1 "Equivalent Transformation Method for Proving Inequalities" Example 14), we get
s2⩽1+9s34s_{2} \leqslant \frac{1 + 9 s_{3}}{4}

Therefore, to prove equation (2), we only need to prove
7729−2592⋅1+9s34+9216s3−32768s32⩾−61+128(1+9s34)2−512s3+464⋅1+9s34⇔781+3384s3−32768s32⩾63+676s3+648s32\begin{array}{l} 7 \sqrt{729 - 2592 \cdot \frac{1 + 9 s_{3}}{4} + 9216 s_{3} - 32768 s_{3}^{2}} \geqslant \\ -61 + 128 \left(\frac{1 + 9 s_{3}}{4}\right)^{2} - 512 s_{3} + 464 \cdot \frac{1 + 9 s_{3}}{4} \Leftrightarrow \\ 7 \sqrt{81 + 3384 s_{3} - 32768 s_{3}^{2}} \geqslant \\ 63 + 676 s_{3} + 648 s_{3}^{2} \end{array}

Squaring both sides of the above equation and rearranging, we get
s3(1260−33504s3−13689s32−6561s33)⩾0⇔s3(1−27s3)(1260+516s3+243)⩾0\begin{array}{l} s_{3} \left(1260 - 33504 s_{3} - 13689 s_{3}^{2} - 6561 s_{3}^{3}\right) \geqslant 0 \Leftrightarrow \\ s_{3} \left(1 - 27 s_{3}\right) \left(1260 + 516 s_{3} + 243\right) \geqslant 0 \end{array}

This inequality is clearly true, therefore equation (2) holds, and thus equation (1) is proven. From the above proof, it is easy to obtain the conditions for equality.

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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.