Track / Stage 7 / 274 of 300 #1674 of 1964
Problem 1674 National olympiad second round; IMO P1/P4 Difficulty 7.7 Prove it
Example 1 (Self-created problem, 1983.01.02) Let x 1 , x 2 , x 3 , x 4 ∈ R x_{1}, x_{2}, x_{3}, x_{4} \in R x 1 , x 2 , x 3 , x 4 ∈ R ∑ i = 1 4 x i = 0 \sum_{i=1}^{4} x_{i}=0 ∑ i = 1 4 x i = 0 ( 1 3 ∑ i = 1 4 x i 2 ) 3 ⩾ ( 1 3 ∑ i = 1 4 x i 3 ) 2 = ( x 2 x 3 x 4 + x 1 x 2 x 4 + x 1 x 3 x 4 + x 1 x 2 x 3 ) 2 \begin{array}{l}
\left(\frac{1}{3} \sum_{i=1}^{4} x_{i}^{2}\right)^{3} \geqslant\left(\frac{1}{3} \sum_{i=1}^{4} x_{i}^{3}\right)^{2}= \\
\left(x_{2} x_{3} x_{4}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{1} x_{2} x_{3}\right)^{2}
\end{array} ( 3 1 ∑ i = 1 4 x i 2 ) 3 ⩾ ( 3 1 ∑ i = 1 4 x i 3 ) 2 = ( x 2 x 3 x 4 + x 1 x 2 x 4 + x 1 x 3 x 4 + x 1 x 2 x 3 ) 2
Equality in (1) holds if and only if three of x 1 , x 2 , x 3 , x 4 x_{1}, x_{2}, x_{3}, x_{4} x 1 , x 2 , x 3 , x 4
This one wants a proof. Work it on paper, then read the official solution and mark
yourself. Be honest about it: the record is only any use to you if it is.
I solved it I didn't Skip
Official solution Prove( ∑ i = 1 4 x i 2 ) 3 = [ x 1 2 + x 2 2 + x 3 2 + ( x 1 + x 2 + x 3 ) 2 ] 3 = [ ( x 2 + x 3 ) 2 + ( x 3 + x 1 ) 2 + ( x 1 + x 2 ) 2 ] 3 ⩾ 27 ( x 2 + x 3 ) 2 ( x 3 + x 1 ) 2 ( x 1 + x 2 ) 2 = 27 [ x 2 x 3 ( x 1 + x 2 + x 3 ) + x 3 x 1 ( x 1 + x 2 + x 3 ) + x 1 x 2 ( x 1 + x 2 + x 3 ) − x 1 x 2 x 3 ] 2 = 27 ( x 2 x 3 x 4 + x 1 x 3 x 4 + x 1 x 2 x 4 + x 1 x 2 x 3 ) 2 \begin{aligned}
\left(\sum_{i=1}^{4} x_{i}^{2}\right)^{3}= & {\left[x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\left(x_{1}+x_{2}+x_{3}\right)^{2}\right]^{3}=} \\
& {\left[\left(x_{2}+x_{3}\right)^{2}+\left(x_{3}+x_{1}\right)^{2}+\left(x_{1}+x_{2}\right)^{2}\right]^{3} \geqslant } \\
& 27\left(x_{2}+x_{3}\right)^{2}\left(x_{3}+x_{1}\right)^{2}\left(x_{1}+x_{2}\right)^{2}= \\
& 27\left[x_{2} x_{3}\left(x_{1}+x_{2}+x_{3}\right)+x_{3} x_{1}\left(x_{1}+x_{2}+x_{3}\right)+\right. \\
& \left.x_{1} x_{2}\left(x_{1}+x_{2}+x_{3}\right)-x_{1} x_{2} x_{3}\right]^{2}= \\
& 27\left(x_{2} x_{3} x_{4}+x_{1} x_{3} x_{4}+x_{1} x_{2} x_{4}+x_{1} x_{2} x_{3}\right)^{2}
\end{aligned} ( i = 1 ∑ 4 x i 2 ) 3 = [ x 1 2 + x 2 2 + x 3 2 + ( x 1 + x 2 + x 3 ) 2 ] 3 = [ ( x 2 + x 3 ) 2 + ( x 3 + x 1 ) 2 + ( x 1 + x 2 ) 2 ] 3 ⩾ 27 ( x 2 + x 3 ) 2 ( x 3 + x 1 ) 2 ( x 1 + x 2 ) 2 = 27 [ x 2 x 3 ( x 1 + x 2 + x 3 ) + x 3 x 1 ( x 1 + x 2 + x 3 ) + x 1 x 2 ( x 1 + x 2 + x 3 ) − x 1 x 2 x 3 ] 2 = 27 ( x 2 x 3 x 4 + x 1 x 3 x 4 + x 1 x 2 x 4 + x 1 x 2 x 3 ) 2
Also27 [ x 2 x 3 ( x 1 + x 2 + x 3 ) + x 3 x 1 ( x 1 + x 2 + x 3 ) + x 1 x 2 ( x 1 + x 2 + x 3 ) − x 1 x 2 x 3 ] = 27 [ x 2 x 3 ( x 2 + x 3 ) + x 3 x 1 ( x 3 + x 1 ) + x 1 x 2 ( x 1 + x 2 ) + 2 x 1 x 2 x 3 ] = 9 [ ( x 2 + x 3 + x 4 ) 3 − x 1 3 − x 2 3 − x 3 3 ] = − 9 ( x 1 3 + x 2 3 + x 3 3 + x 4 3 ) \begin{array}{l}
27\left[x_{2} x_{3}\left(x_{1}+x_{2}+x_{3}\right)+x_{3} x_{1}\left(x_{1}+x_{2}+x_{3}\right)+x_{1} x_{2}\left(x_{1}+x_{2}+x_{3}\right)-x_{1} x_{2} x_{3}\right]= \\
27\left[x_{2} x_{3}\left(x_{2}+x_{3}\right)+x_{3} x_{1}\left(x_{3}+x_{1}\right)+x_{1} x_{2}\left(x_{1}+x_{2}\right)+2 x_{1} x_{2} x_{3}\right]= \\
9\left[\left(x_{2}+x_{3}+x_{4}\right)^{3}-x_{1}^{3}-x_{2}^{3}-x_{3}^{3}\right]= \\
-9\left(x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}\right)
\end{array} 27 [ x 2 x 3 ( x 1 + x 2 + x 3 ) + x 3 x 1 ( x 1 + x 2 + x 3 ) + x 1 x 2 ( x 1 + x 2 + x 3 ) − x 1 x 2 x 3 ] = 27 [ x 2 x 3 ( x 2 + x 3 ) + x 3 x 1 ( x 3 + x 1 ) + x 1 x 2 ( x 1 + x 2 ) + 2 x 1 x 2 x 3 ] = 9 [ ( x 2 + x 3 + x 4 ) 3 − x 1 3 − x 2 3 − x 3 3 ] = − 9 ( x 1 3 + x 2 3 + x 3 3 + x 4 3 )
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Source: NuminaMath-1.5 ,
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