Maths Olympiad Prep

Library / /219 of 520

Algebra Difficulty 6.5 National olympiad Prove it

Example 5.85.8 Given a,b,c0,a3+b3+c3=3a, b, c \geqslant 0, a^{3}+b^{3}+c^{3}=3, prove that
a4b4+b4c4+c4a43a^{4} b^{4}+b^{4} c^{4}+c^{4} a^{4} \leqslant 3

Solution

Prove that if we directly homogenize the degree, it would be too high and difficult to handle. Consider using the conditions to gradually "increase the degree".
9b4c4=9b3c3bc3b3c3(b3+c3+1)=3b3c3(b3+c3)+3b3c3=b3c3(b3+c3)+2b3c3(b3+c3)+3b3c3a9+3a3b3c3+2b3c3(b3+c3)+3b3c3=(3rd degree Schur inequality)3(a3)2=27\begin{aligned} 9 \sum b^{4} c^{4}= & 9 \sum b^{3} c^{3} \cdot b c \leqslant 3 \sum b^{3} c^{3}\left(b^{3}+c^{3}+1\right)= \\ & 3 \sum b^{3} c^{3}\left(b^{3}+c^{3}\right)+3 \sum b^{3} c^{3}= \\ & \sum b^{3} c^{3}\left(b^{3}+c^{3}\right)+2 \sum b^{3} c^{3}\left(b^{3}+c^{3}\right)+3 \sum b^{3} c^{3} \leqslant \\ & \sum a^{9}+3 a^{3} b^{3} c^{3}+2 \sum b^{3} c^{3}\left(b^{3}+c^{3}\right)+ \\ & 3 \sum b^{3} c^{3}=(3 \text{rd degree Schur inequality}) \\ & 3\left(\sum a^{3}\right)^{2}=27 \end{aligned}

Thus, the proposition is proved!

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.